MATH 114 - QUANTITATIVE REASONING -
Solving Linear And Quadratic Equations And
Inequalities Question Bank
Question 1
Solve the following linear equation:
3x−5=2x+ 7
Solution:
To solve the linear equation 3x−5=2x+ 7, we need to isolate xon one
side of the equation.
3x−5=2x+ 7
3x−2x= 7 + 5 (Add 2xto both sides)
x= 12 (Simplify)
Therefore, the solution to the equation is x= 12.Question 1.
Solve the following linear equation:
3x−5=2x+ 7
Solution:
To solve the linear equation 3x−5=2x+ 7, we need to isolate xon
one side of the equation.
3x−5=2x+ 7
3x−2x= 7 + 5 (Add 2xto both sides)
x= 12 (Simplify)
Therefore, the solution to the equation is x= 12.
1
Question 2
Solve the following linear equation for x:
3(2x−4) = 5x+ 7
Solution:
Step 1: Distribute the 3 on the left side of the equation.
3(2x−4) = 5x+ 7
6x−12 = 5x+ 7
Step 2: Move all terms involving xto one side of the equation and
constants to the other side.
6x−12 = 5x+ 7
6x−5x= 7 + 12
x= 19
Therefore, the solution to the equation 3(2x−4) = 5x+ 7 is x=
19.Question 2:
Solve the following linear equation for x:
3(2x−4) = 5x+ 7
Solution:
Step 1: Distribute the 3 on the left side of the equation.
3(2x−4) = 5x+ 7
6x−12 = 5x+ 7
Step 2: Move all terms involving xto one side of the equation and
constants to the other side.
6x−12 = 5x+ 7
6x−5x= 7 + 12
x= 19
Therefore, the solution to the equation 3(2x−4) = 5x+ 7 is x= 19.
Question 3
Solution: To solve the quadratic equation 2x2+ 5x−3=0, we can
use the quadratic formula: The quadratic formula is given by:
x=−b±√b2−4ac
2a
2
Given the quadratic equation 2x2+ 5x−3=0, we have: a= 2,b= 5,
and c=−3
Plugging these values into the quadratic formula, we get:
x=−5±p52−4(2)(−3)
2(2)
x=−5±√25 + 24
4
x=−5±√49
4
x=−5±7
4
Therefore, the solutions to the quadratic equation 2x2+ 5x−3=0
are:
x=−5+7
4=2
4=1
2
and
x=−5−7
4=−12
4=−3
So, the solutions to the quadratic equation 2x2+ 5x−3=0are x=
1
2and x=−3.Question 3: Solve the following quadratic equation:
2x2+ 5x−3=0
Solution: To solve the quadratic equation 2x2+ 5x−3=0, we can
use the quadratic formula: The quadratic formula is given by:
x=−b±√b2−4ac
2a
Given the quadratic equation 2x2+ 5x−3=0, we have: a= 2,b= 5,
and c=−3
Plugging these values into the quadratic formula, we get:
x=−5±p52−4(2)(−3)
2(2)
x=−5±√25 + 24
4
x=−5±√49
4
x=−5±7
4
3
Therefore, the solutions to the quadratic equation 2x2+ 5x−3=0
are:
x=−5+7
4=2
4=1
2
and
x=−5−7
4=−12
4=−3
So, the solutions to the quadratic equation 2x2+ 5x−3 = 0 are x=1
2
and x=−3.
Question 4
Solution: To solve the quadratic equation x2+ 6x+ 8 = 0, we can
use the quadratic formula:
x=−b±√b2−4ac
2a
where a= 1,b= 6, and c= 8 in this case. Plug in these values into
the formula:
x=−6±√62−4·1·8
2·1
x=−6±√36 −32
2
x=−6±√4
2
Now, calculate the two possible solutions for xby splitting the ±:
x1=−6+2
2=−2
x2=−6−2
2=−4
Therefore, the solutions to the quadratic equation x2+6x+8 = 0 are
x=−2and x=−4.Question 4: Solve the following quadratic equation
for x:
x2+ 6x+ 8 = 0
Solution: To solve the quadratic equation x2+ 6x+ 8 = 0, we can
use the quadratic formula:
x=−b±√b2−4ac
2a
4
where a= 1,b= 6, and c= 8 in this case. Plug in these values into
the formula:
x=−6±√62−4·1·8
2·1
x=−6±√36 −32
2
x=−6±√4
2
Now, calculate the two possible solutions for xby splitting the ±:
x1=−6+2
2=−2
x2=−6−2
2=−4
Therefore, the solutions to the quadratic equation x2+ 6x+ 8 = 0
are x=−2and x=−4.
Question 5
Step-by-step solution: Step 1: Identify the coefficients of the
quadratic equation 2x2−5x−3=0. In this case, a= 2,b=−5, and
c=−3. Step 2: Use the quadratic formula x=−b±√b2−4ac
2ato solve
for x. Step 3: Substitute the values of a,b, and cinto the quadratic for-
mula: x=−(−5) ±p(−5)2−4·2·(−3)
2·2. Step 4: Simplify the equation:
x=5±√25 + 24
4. Step 5: Further simplify the equation: x=5±√49
4.
Step 6: Calculate the square root of 49: x=5±7
4. Step 7: Determine
the two possible solutions: - Solution 1: x=5+7
4=12
4= 3. - Solution
2: x=5−7
4=−2
4=−1
2. Therefore, the solutions to the quadratic
equation 2x2−5x−3=0are x= 3 and x=−1
2.Question 5: Solve the
following quadratic equation for x:2x2−5x−3=0.
Step-by-step solution: Step 1: Identify the coefficients of the
quadratic equation 2x2−5x−3=0. In this case, a= 2,b=−5, and
c=−3. Step 2: Use the quadratic formula x=−b±√b2−4ac
2ato solve
for x. Step 3: Substitute the values of a,b, and cinto the quadratic for-
mula: x=−(−5) ±p(−5)2−4·2·(−3)
2·2. Step 4: Simplify the equation:
5
x=5±√25 + 24
4. Step 5: Further simplify the equation: x=5±√49
4.
Step 6: Calculate the square root of 49: x=5±7
4. Step 7: Determine
the two possible solutions: - Solution 1: x=5+7
4=12
4= 3. - Solution
2: x=5−7
4=−2
4=−1
2. Therefore, the solutions to the quadratic
equation 2x2−5x−3=0are x= 3 and x=−1
2.
Question 6
Step-by-step solution: To solve the quadratic inequality 4x2−12x+
9≥0, we first find the roots of the associated quadratic equation
4x2−12x+ 9 = 0.
Step 1: Start by factoring the quadratic equation:
4x2−12x+ 9 = 0
(2x−3)2= 0
2x−3=0
2x= 3
x=3
2
Thus, the quadratic equation has one repeated root at x=3
2.
Step 2: Plot the critical point on the number line, x=3
2.
Step 3: Test a point in each of the intervals created by the critical
point (e.g., x= 0,x= 1) in the original inequality to determine the
sign of the expression: For x= 0:4(0)2−12(0) + 9 = 9 Since 9≥0, the
inequality holds for this interval.
For x= 1:4(1)2−12(1) + 9 = 1 Since 1≥0, the inequality holds for
this interval.
Step 4: Based on the tests, the solution to the quadratic inequality
4x2−12x+ 9 ≥0is x≤3
2.Question 6: Solve the following quadratic
inequality: 4x2−12x+ 9 ≥0
Step-by-step solution: To solve the quadratic inequality 4x2−12x+
9≥0, we first find the roots of the associated quadratic equation
4x2−12x+ 9 = 0.
Step 1: Start by factoring the quadratic equation:
4x2−12x+ 9 = 0
(2x−3)2= 0
2x−3=0
6
2x= 3
x=3
2
Thus, the quadratic equation has one repeated root at x=3
2.
Step 2: Plot the critical point on the number line, x=3
2.
Step 3: Test a point in each of the intervals created by the critical
point (e.g., x= 0,x= 1) in the original inequality to determine the
sign of the expression: For x= 0:4(0)2−12(0) + 9 = 9 Since 9≥0, the
inequality holds for this interval.
For x= 1:4(1)2−12(1) + 9 = 1 Since 1≥0, the inequality holds for
this interval.
Step 4: Based on the tests, the solution to the quadratic inequality
4x2−12x+ 9 ≥0is x≤3
2.
Question 7
Step-by-step solution: 1. First, factor the quadratic expression to
find the roots of the related equation 2x2−5x+ 2 = 0. The factored
form is (2x−1)(x−2) = 0, which gives the roots x=1
2and x= 2. 2.
Plot these roots on a number line to divide it into three intervals:
(−∞,1
2),(1
2,2), and (2,∞). 3. Choose a test point from each interval
to determine the sign of the inequality in that interval. For example,
for the interval (−∞,1
2), choose x= 0 as a test point. 4. Substitute
the test points back into the inequality 2x2−5x+ 2 >0and check the
sign of the expression to determine where it is positive. 5. Based on
the signs of these test points, we find that the inequality is satisfied
for x∈(1
2,2). 6. Therefore, the solution to the quadratic inequality
2x2−5x+ 2 >0is x∈(1
2,2).Question 7: Solve the quadratic inequality
2x2−5x+ 2 >0
Step-by-step solution: 1. First, factor the quadratic expression to
find the roots of the related equation 2x2−5x+ 2 = 0. The factored
form is (2x−1)(x−2) = 0, which gives the roots x=1
2and x= 2. 2.
Plot these roots on a number line to divide it into three intervals:
(−∞,1
2),(1
2,2), and (2,∞). 3. Choose a test point from each interval
to determine the sign of the inequality in that interval. For example,
for the interval (−∞,1
2), choose x= 0 as a test point. 4. Substitute
the test points back into the inequality 2x2−5x+ 2 >0and check the
sign of the expression to determine where it is positive. 5. Based on
the signs of these test points, we find that the inequality is satisfied
for x∈(1
2,2). 6. Therefore, the solution to the quadratic inequality
2x2−5x+ 2 >0is x∈(1
2,2).
7
Question 8
Step-by-step Solution: 1. Begin by finding the critical points of
the inequality by setting the quadratic expression equal to zero: 2x2−
3x−2=0. 2. To solve the quadratic equation 2x2−3x−2=0, use the
quadratic formula: x=−b±√b2−4ac
2a, where a= 2,b=−3, and c=−2.
3. Calculate the discriminant, b2−4ac for the quadratic formula:
(−3)2−4∗2∗(−2) = 9 + 16 = 25. 4. Since the discriminant is positive
(25 >0), the quadratic equation has two real roots. 5. Substitute
the values of a,b, and the discriminant into the quadratic formula
to find the two roots of the quadratic equation. 6. Once you have
found the roots, x1and x2, plot them on a number line to create
intervals. 7. Test each interval with a test point to determine the
sign of the quadratic expression in that interval. 8. The solution to
the inequality 2x2−3x−2<0will be the intervals where the sign of
the expression is negative. 9. Write the final solution as an interval
or union of intervals.Question 8: Solve the quadratic inequality 2x2−
3x−2<0.
Step-by-step Solution: 1. Begin by finding the critical points of
the inequality by setting the quadratic expression equal to zero: 2x2−
3x−2=0. 2. To solve the quadratic equation 2x2−3x−2=0, use the
quadratic formula: x=−b±√b2−4ac
2a, where a= 2,b=−3, and c=−2.
3. Calculate the discriminant, b2−4ac for the quadratic formula:
(−3)2−4∗2∗(−2) = 9 + 16 = 25. 4. Since the discriminant is positive
(25 >0), the quadratic equation has two real roots. 5. Substitute
the values of a,b, and the discriminant into the quadratic formula
to find the two roots of the quadratic equation. 6. Once you have
found the roots, x1and x2, plot them on a number line to create
intervals. 7. Test each interval with a test point to determine the
sign of the quadratic expression in that interval. 8. The solution to
the inequality 2x2−3x−2<0will be the intervals where the sign of
the expression is negative. 9. Write the final solution as an interval
or union of intervals.
Question 9
Solve the quadratic inequality: 2x2+ 5x−3<0.
Solution:
To solve the quadratic inequality 2x2+ 5x−3<0, we need to find
the values of xthat satisfy the inequality.
Step 1: Find the roots of the quadratic equation 2x2+ 5x−3=0
by factoring or using the quadratic formula.
The quadratic equation 2x2+ 5x−3=0can be factored as (2x−
1)(x+ 3) = 0.
8
Setting each factor to zero gives 2x−1 = 0 and x+ 3 = 0, which
leads to x=1
2and x=−3.
So, the roots of the quadratic equation are x=1
2and x=−3.
Step 2: Plot these roots on a number line to create intervals.
−3divides the number line into two intervals: (−∞,−3) and (−3,1
2).
Step 3: Test points within each interval to determine the sign of
the quadratic expression.
For x=−4(in (−∞,−3)):
Plugging x=−4into 2x2+5x−3gives 2(−4)2+5(−4)−3 = 32−20−3 =
9>0.
For x= 0 (in (−3,1
2)):
Plugging x= 0 into 2x2+ 5x−3gives 2(0)2+ 5(0) −3 = −3<0.
Step 4: Analyze the signs in the intervals based on the test points.
Since the inequality 2x2+ 5x−3<0is satisfied for x∈(−3,1
2), the
solution to the inequality is (−3,1
2).Question 9:
Solve the quadratic inequality: 2x2+ 5x−3<0.
Solution:
To solve the quadratic inequality 2x2+ 5x−3<0, we need to find
the values of xthat satisfy the inequality.
Step 1: Find the roots of the quadratic equation 2x2+ 5x−3=0
by factoring or using the quadratic formula.
The quadratic equation 2x2+ 5x−3=0can be factored as (2x−
1)(x+ 3) = 0.
Setting each factor to zero gives 2x−1 = 0 and x+ 3 = 0, which
leads to x=1
2and x=−3.
So, the roots of the quadratic equation are x=1
2and x=−3.
Step 2: Plot these roots on a number line to create intervals.
−3divides the number line into two intervals: (−∞,−3) and (−3,1
2).
Step 3: Test points within each interval to determine the sign of
the quadratic expression.
For x=−4(in (−∞,−3)):
Plugging x=−4into 2x2+5x−3gives 2(−4)2+5(−4)−3 = 32−20−3 =
9>0.
For x= 0 (in (−3,1
2)):
Plugging x= 0 into 2x2+ 5x−3gives 2(0)2+ 5(0) −3 = −3<0.
Step 4: Analyze the signs in the intervals based on the test points.
Since the inequality 2x2+ 5x−3<0is satisfied for x∈(−3,1
2), the
solution to the inequality is (−3,1
2).
Question 10
Step-by-step solution: 1. Add 5 to both sides of the inequality:
9
2x−5 + 5 >3 + 5 2. Simplify: 2x > 83. Divide by 2 on both sides to
isolate the variable: 2x
2>8
24. Solve for x:x > 4
Therefore, the solution to the inequality 2x−5>3is x > 4.Question
10: Solve the linear inequality: 2x−5>3
Step-by-step solution: 1. Add 5 to both sides of the inequality:
2x−5 + 5 >3 + 5 2. Simplify: 2x > 83. Divide by 2 on both sides to
isolate the variable: 2x
2>8
24. Solve for x:x > 4
Therefore, the solution to the inequality 2x−5>3is x > 4.
Question 11
Step-by-step solution: 1. Start by factoring the quadratic expres-
sion 3x2−4x > 0:
3x2−4x > 0
x(3x−4) >0
2. The critical points are when x= 0 and when 3x−4=0, which
gives x=4
3. 3. Plot these critical points on a number line and test
intervals to determine the solutions. 4. Test the interval (−,0) with
a test point x=−1:
(−1)(3(−1) −4) >0
−3+4>0
1>0
5. Test the interval (0,4
3)with a test point x= 1:
1(3(1) −4) >0
3−4<0
−1<0
6. Test the interval (4
3,)with a test point x= 2:
2(3(2) −4) >0
2(6 −4) >0
4>0
7. Therefore, the solution to the inequality 3x2−4x > 0is x < 0or
x > 4
3.Question 11: Solve the quadratic inequality 3x2−4x > 0.
Step-by-step solution: 1. Start by factoring the quadratic expres-
sion 3x2−4x > 0:
3x2−4x > 0
x(3x−4) >0
10
2. The critical points are when x= 0 and when 3x−4=0, which
gives x=4
3. 3. Plot these critical points on a number line and test
intervals to determine the solutions. 4. Test the interval (−,0) with
a test point x=−1:
(−1)(3(−1) −4) >0
−3+4>0
1>0
5. Test the interval (0,4
3)with a test point x= 1:
1(3(1) −4) >0
3−4<0
−1<0
6. Test the interval (4
3,)with a test point x= 2:
2(3(2) −4) >0
2(6 −4) >0
4>0
7. Therefore, the solution to the inequality 3x2−4x > 0is x < 0or
x > 4
3.
Question 12
Step-by-step solution: To solve the inequality 2x2+ 5x−3≤0, we
follow these steps:
1. Factor the quadratic expression: 2x2+ 5x−3 = (2x−1)(x+ 3).
2. Set each factor equal to zero and solve for x:
2x−1=0
2x= 1
x=1
2
and
x+ 3 = 0
x=−3
3. Mark the critical points x=−3and x=1
2on a number line.
4. Test each interval created by the critical points with a test value
(e.g., 0): - For x < −3: Test x=−4:(2(−4) −1)(−4 + 3) = (−9)(−1) >0.
11
- For −3<x< 1
2: Test x= 0:(2(0) −1)(0 + 3) = (−1)(3) <0. - For x > 1
2:
Test x= 1:(2(1) −1)(1 + 3) = (1)(4) >0.
5. Determine the solution: The inequality 2x2+ 5x−3≤0is true
for −∞ < x ≤ −3and 1
2≥x < ∞.
Therefore, the solution to the inequality is x∈(−∞,−3]∪1
2,∞.Question
12: Solve the quadratic inequality 2x2+ 5x−3≤0.
Step-by-step solution: To solve the inequality 2x2+ 5x−3≤0, we
follow these steps:
1. Factor the quadratic expression: 2x2+ 5x−3 = (2x−1)(x+ 3).
2. Set each factor equal to zero and solve for x:
2x−1=0
2x= 1
x=1
2
and
x+ 3 = 0
x=−3
3. Mark the critical points x=−3and x=1
2on a number line.
4. Test each interval created by the critical points with a test value
(e.g., 0): - For x < −3: Test x=−4:(2(−4) −1)(−4 + 3) = (−9)(−1) >0.
- For −3<x< 1
2: Test x= 0:(2(0) −1)(0 + 3) = (−1)(3) <0. - For x > 1
2:
Test x= 1:(2(1) −1)(1 + 3) = (1)(4) >0.
5. Determine the solution: The inequality 2x2+ 5x−3≤0is true
for −∞ < x ≤ −3and 1
2≥x < ∞.
Therefore, the solution to the inequality is x∈(−∞,−3] ∪1
2,∞.
Question 13
Solve the inequality:
2x−5<3(x+ 4)
Step-by-step Solution:
Given inequality:
2x−5<3(x+ 4)
Expand the right side:
2x−5<3x+ 12
Subtract 2xfrom both sides:
12
−5< x + 12
Subtract 12 from both sides:
−17 < x
Therefore, the solution to the inequality is x > −17.Question 13:
Solve the inequality:
2x−5<3(x+ 4)
Step-by-step Solution:
Given inequality:
2x−5<3(x+ 4)
Expand the right side:
2x−5<3x+ 12
Subtract 2xfrom both sides:
−5< x + 12
Subtract 12 from both sides:
−17 < x
Therefore, the solution to the inequality is x > −17.
Question 14
Step-by-step solution: 1. Rewrite the inequality in standard form:
2x2−5x−3<0. 2. Factor the quadratic expression: (2x+ 1)(x−3) <0.
3. Determine the critical points by setting each factor equal to zero:
2x+ 1 = 0 and x−3=0. 4. Solve for x:2x=−1⇒x=−1
2and x= 3.
5. Plot the critical points on a number line and test intervals created
by these points in the original inequality. 6. Test interval 1: Choose
a test point x= 0. Substitute into the expression: 2(0)2−5(0) −3<0,
which simplifies to −3<0, and is true. Therefore, interval 1 is part of
the solution. 7. Test interval 2: Choose a test point x= 4. Substitute
into the expression: 2(4)2−5(4) −3<0, which simplifies to 17 <0,
and is false. Therefore, interval 2 is not part of the solution. 8. The
solution to the quadratic inequality is −1
2< x < 3.Question 14: Solve
the quadratic inequality 2x2−5x < 3.
Step-by-step solution: 1. Rewrite the inequality in standard form:
2x2−5x−3<0. 2. Factor the quadratic expression: (2x+ 1)(x−3) <0.
3. Determine the critical points by setting each factor equal to zero:
13
2x+ 1 = 0 and x−3=0. 4. Solve for x:2x=−1⇒x=−1
2and x= 3.
5. Plot the critical points on a number line and test intervals created
by these points in the original inequality. 6. Test interval 1: Choose
a test point x= 0. Substitute into the expression: 2(0)2−5(0) −3<0,
which simplifies to −3<0, and is true. Therefore, interval 1 is part of
the solution. 7. Test interval 2: Choose a test point x= 4. Substitute
into the expression: 2(4)2−5(4) −3<0, which simplifies to 17 <0,
and is false. Therefore, interval 2 is not part of the solution. 8. The
solution to the quadratic inequality is −1
2<x<3.
Question 15
3x2−5x > 2
Step-by-step solution:
1. Start by moving all terms to one side of the inequality to set it
equal to zero:
3x2−5x−2>0
2. Next, factor the quadratic expression:
3x2−5x−2 = (3x+ 1)(x−2) >0
3. Find the critical points by setting each factor equal to zero:
3x+ 1 = 0 →x=−1
3
x−2=0→x= 2
4. Create a number line and plot the critical points on it:
−1
32
−0 +
5. Test each interval on the number line by choosing a test point
in each interval and checking its sign when substituted into the in-
equality:
For x < −1
3: Choose x=−1
3(−1)2−5(−1) −2=8>0
So, this interval is positive.
For −1
3<x<2: Choose x= 0
3(0)2−5(0) −2 = −2<0
So, this interval is negative.
For x > 2: Choose x= 3
3(3)2−5(3) −2 = 16 >0
So, this interval is positive.
6. Determine the solution by analyzing the signs on the number
line:
The solution to the inequality 3x2−5x−2>0is −1
3<x<2, which
can be written in interval notation as (−1
3,2).Question 15: Solve the
following quadratic inequality:
3x2−5x > 2
Step-by-step solution:
14
1. Start by moving all terms to one side of the inequality to set it
equal to zero:
3x2−5x−2>0
2. Next, factor the quadratic expression:
3x2−5x−2 = (3x+ 1)(x−2) >0
3. Find the critical points by setting each factor equal to zero:
3x+ 1 = 0 →x=−1
3
x−2=0→x= 2
4. Create a number line and plot the critical points on it:
−1
32
−0 +
5. Test each interval on the number line by choosing a test point
in each interval and checking its sign when substituted into the in-
equality:
For x < −1
3: Choose x=−1
3(−1)2−5(−1) −2=8>0
So, this interval is positive.
For −1
3<x<2: Choose x= 0
3(0)2−5(0) −2 = −2<0
So, this interval is negative.
For x > 2: Choose x= 3
3(3)2−5(3) −2 = 16 >0
So, this interval is positive.
6. Determine the solution by analyzing the signs on the number
line:
The solution to the inequality 3x2−5x−2>0is −1
3<x<2, which
can be written in interval notation as (−1
3,2).
Question 16
Step-by-step solution: 1. Start by finding the roots of the corre-
sponding quadratic equation 2x2−7x+ 3 = 0. 2. Use the quadratic
formula x=−b±√b2−4ac
2awhere a= 2,b=−7, and c= 3 to find the roots.
3. Calculate the discriminant ∆ = b2−4ac to determine the nature
of the roots. 4. If the discriminant is positive, there are two distinct
real roots. If it is zero, there is one real root. If it is negative, there
are no real roots. 5. Once you have found the roots, plot them on a
number line to divide the number line into intervals. 6. Test a value
from each interval in the original inequality to determine the solution
set. 7. The solution to the quadratic inequality 2x2−7x+ 3 <0will
be the values that make the inequality true.Question 16: Solve the
quadratic inequality 2x2−7x+ 3 <0.
Step-by-step solution: 1. Start by finding the roots of the corre-
sponding quadratic equation 2x2−7x+ 3 = 0. 2. Use the quadratic
15
formula x=−b±√b2−4ac
2awhere a= 2,b=−7, and c= 3 to find the roots.
3. Calculate the discriminant ∆ = b2−4ac to determine the nature
of the roots. 4. If the discriminant is positive, there are two distinct
real roots. If it is zero, there is one real root. If it is negative, there
are no real roots. 5. Once you have found the roots, plot them on a
number line to divide the number line into intervals. 6. Test a value
from each interval in the original inequality to determine the solution
set. 7. The solution to the quadratic inequality 2x2−7x+ 3 <0will
be the values that make the inequality true.
Question 17
Step-by-step solution:
3
2x−1≤1
2(4x+ 7)
3x−2≤2x+7
2(Distribute on the right side)
3x−2x≤7
2+ 2 (Subtract 2xfrom both sides)
x≤11
2
Therefore, the solution to the inequality is x≤11
2.Question 17:
Solve the inequality 3
2x−1≤1
2(4x+ 7).
Step-by-step solution:
3
2x−1≤1
2(4x+ 7)
3x−2≤2x+7
2(Distribute on the right side)
3x−2x≤7
2+ 2 (Subtract 2xfrom both sides)
x≤11
2
Therefore, the solution to the inequality is x≤11
2.
Question 18
Solve the following quadratic inequality:
(x−3)(x+ 2) ≥0
Solution:
16
To solve the quadratic inequality (x−3)(x+ 2) ≥0, we need to find
the critical points where the expression equals zero and determine
the sign of the expression in each interval.
1. Find the critical points by setting the expression equal to zero:
(x−3)(x+ 2) = 0
This gives us x= 3 and x=−2as critical points.
2. Plot these critical points on a number line:
−2 3
3. Test the intervals created by the critical points by choosing test
points and determining the sign of the expression in each interval.
- For x < −2, choose x=−3:
(−3−3)(−3 + 2) = (−6)(−1) = 6 >0
Since the expression is positive, solutions in this interval are included.
- For −2<x<3, choose x= 0:
(0 −3)(0 + 2) = (−3)(2) = −6<0
Since the expression is negative, solutions in this interval are not
included.
- For x > 3, choose x= 4:
(4 −3)(4 + 2) = (1)(6) = 6 >0
Since the expression is positive, solutions in this interval are included.
4. Combine the results to write the solution set:
x∈(−∞,−2] ∪[3,∞)
Therefore, the solution to the inequality (x−3)(x+ 2) ≥0is x∈
(−∞,−2] ∪[3,∞).Question 18:
Solve the following quadratic inequality:
(x−3)(x+ 2) ≥0
Solution:
To solve the quadratic inequality (x−3)(x+ 2) ≥0, we need to find
the critical points where the expression equals zero and determine
the sign of the expression in each interval.
1. Find the critical points by setting the expression equal to zero:
(x−3)(x+ 2) = 0
This gives us x= 3 and x=−2as critical points.
17
2. Plot these critical points on a number line:
−2 3
3. Test the intervals created by the critical points by choosing test
points and determining the sign of the expression in each interval.
- For x < −2, choose x=−3:
(−3−3)(−3 + 2) = (−6)(−1) = 6 >0
Since the expression is positive, solutions in this interval are included.
- For −2<x<3, choose x= 0:
(0 −3)(0 + 2) = (−3)(2) = −6<0
Since the expression is negative, solutions in this interval are not
included.
- For x > 3, choose x= 4:
(4 −3)(4 + 2) = (1)(6) = 6 >0
Since the expression is positive, solutions in this interval are included.
4. Combine the results to write the solution set:
x∈(−∞,−2] ∪[3,∞)
Therefore, the solution to the inequality (x−3)(x+ 2) ≥0is x∈
(−∞,−2] ∪[3,∞).
Question 19
Solve the following quadratic inequality:
2x2−5x+ 2 <0
Step-by-Step Solution:
To solve the given quadratic inequality, we first rewrite it in the
standard form:
2x2−5x+ 2 <0
Next, we factor the quadratic expression:
2x2−4x−x+ 2 <0
2x(x−2) −1(x−2) <0
(2x−1)(x−2) <0
18
Now, we find the critical points by setting each factor equal to
zero:
2x−1=0⇒x=1
2
x−2 = 0 ⇒x= 2
These critical points divide the real number line into three inter-
vals: (−∞,1
2),(1
2,2),(2,+∞).
We test each interval by choosing a test point and plugging it into
the inequality to determine the sign of the expression:
For x= 0:(2(0) −1)(0 −2) = (−1)(−2) = 2 >0, which is false.
For x= 1:(2(1) −1)(1 −2) = (1)(−1) = −1<0, which is true.
For x= 3:(2(3) −1)(3 −2) = (5)(1) = 5 >0, which is false.
Therefore, the solution to the inequality 2x2−5x+ 2 <0is:
x∈1
2,2
Question 19:
Solve the following quadratic inequality:
2x2−5x+ 2 <0
Step-by-Step Solution:
To solve the given quadratic inequality, we first rewrite it in the
standard form:
2x2−5x+ 2 <0
Next, we factor the quadratic expression:
2x2−4x−x+ 2 <0
2x(x−2) −1(x−2) <0
(2x−1)(x−2) <0
Now, we find the critical points by setting each factor equal to
zero:
2x−1=0⇒x=1
2
x−2 = 0 ⇒x= 2
These critical points divide the real number line into three inter-
vals: (−∞,1
2),(1
2,2),(2,+∞).
We test each interval by choosing a test point and plugging it into
the inequality to determine the sign of the expression:
For x= 0:(2(0) −1)(0 −2) = (−1)(−2) = 2 >0, which is false.
For x= 1:(2(1) −1)(1 −2) = (1)(−1) = −1<0, which is true.
For x= 3:(2(3) −1)(3 −2) = (5)(1) = 5 >0, which is false.
19
Therefore, the solution to the inequality 2x2−5x+ 2 <0is:
x∈1
2,2
Question 20
3x2+ 4x−4>0
Step-by-step solution: 1. Begin by rewriting the inequality in
standard form:
3x2+ 4x−4>0
2. Identify the coefficients of the quadratic equation:
a= 3, b = 4, c =−4
3. Calculate the discriminant:
∆ = b2−4ac
∆=42−4(3)(−4)
∆ = 16 + 48
∆ = 64
4. Determine the nature of the roots using the discriminant: - If
∆>0, the inequality will have two distinct real roots. - If ∆=0, the
inequality will have a repeated real root. - If ∆<0, the inequality
will have no real roots.
5. Since ∆>0, the inequality will have two distinct real roots. To
find the roots, use the quadratic formula:
x=−b±√∆
2a
x=−4±√64
2(3)
x=−4±8
6
6. We get two roots:
x1=−4+8
6=4
6=2
3
x2=−4−8
6=−12
6=−2
20
7. Now, we need to determine the intervals where the inequality is
satisfied. To do this, we create a sign chart using the roots we found.
x < −2−2<x< 2
3x > 2
3
3x2+ 4x−4−+ +
8. Since the inequality is greater than zero (>0), the solution is
where the quadratic expression is positive:
x∈(−∞,−2) ∪2
3,∞
Therefore, the solution set for the quadratic inequality 3x2+ 4x−
4>0is (−∞,−2) ∪2
3,∞.Question 20: Solve the following quadratic
inequality:
3x2+ 4x−4>0
Step-by-step solution: 1. Begin by rewriting the inequality in
standard form:
3x2+ 4x−4>0
2. Identify the coefficients of the quadratic equation:
a= 3, b = 4, c =−4
3. Calculate the discriminant:
∆ = b2−4ac
∆=42−4(3)(−4)
∆ = 16 + 48
∆ = 64
4. Determine the nature of the roots using the discriminant: - If
∆>0, the inequality will have two distinct real roots. - If ∆=0, the
inequality will have a repeated real root. - If ∆<0, the inequality
will have no real roots.
5. Since ∆>0, the inequality will have two distinct real roots. To
find the roots, use the quadratic formula:
x=−b±√∆
2a
x=−4±√64
2(3)
x=−4±8
6
21
Question 2
Solve the following linear equation for x:
3(2x−4) = 5x+ 7
Solution:
Step 1: Distribute the 3 on the left side of the equation.
3(2x−4) = 5x+ 7
6x−12 = 5x+ 7
Step 2: Move all terms involving xto one side of the equation and
constants to the other side.
6x−12 = 5x+ 7
6x−5x= 7 + 12
x= 19
Therefore, the solution to the equation 3(2x−4) = 5x+ 7 is x=
19.Question 2:
Solve the following linear equation for x:
3(2x−4) = 5x+ 7
Solution:
Step 1: Distribute the 3 on the left side of the equation.
3(2x−4) = 5x+ 7
6x−12 = 5x+ 7
Step 2: Move all terms involving xto one side of the equation and
constants to the other side.
6x−12 = 5x+ 7
6x−5x= 7 + 12
x= 19
Therefore, the solution to the equation 3(2x−4) = 5x+ 7 is x= 19.
Question 3
Solution: To solve the quadratic equation 2x2+ 5x−3=0, we can
use the quadratic formula: The quadratic formula is given by:
x=−b±√b2−4ac
2a
2
Given the quadratic equation 2x2+ 5x−3=0, we have: a= 2,b= 5,
and c=−3
Plugging these values into the quadratic formula, we get:
x=−5±p52−4(2)(−3)
2(2)
x=−5±√25 + 24
4
x=−5±√49
4
x=−5±7
4
Therefore, the solutions to the quadratic equation 2x2+ 5x−3=0
are:
x=−5+7
4=2
4=1
2
and
x=−5−7
4=−12
4=−3
So, the solutions to the quadratic equation 2x2+ 5x−3=0are x=
1
2and x=−3.Question 3: Solve the following quadratic equation:
2x2+ 5x−3=0
Solution: To solve the quadratic equation 2x2+ 5x−3=0, we can
use the quadratic formula: The quadratic formula is given by:
x=−b±√b2−4ac
2a
Given the quadratic equation 2x2+ 5x−3=0, we have: a= 2,b= 5,
and c=−3
Plugging these values into the quadratic formula, we get:
x=−5±p52−4(2)(−3)
2(2)
x=−5±√25 + 24
4
x=−5±√49
4
x=−5±7
4
3
Therefore, the solutions to the quadratic equation 2x2+ 5x−3=0
are:
x=−5+7
4=2
4=1
2
and
x=−5−7
4=−12
4=−3
So, the solutions to the quadratic equation 2x2+ 5x−3 = 0 are x=1
2
and x=−3.
Question 4
Solution: To solve the quadratic equation x2+ 6x+ 8 = 0, we can
use the quadratic formula:
x=−b±√b2−4ac
2a
where a= 1,b= 6, and c= 8 in this case. Plug in these values into
the formula:
x=−6±√62−4·1·8
2·1
x=−6±√36 −32
2
x=−6±√4
2
Now, calculate the two possible solutions for xby splitting the ±:
x1=−6+2
2=−2
x2=−6−2
2=−4
Therefore, the solutions to the quadratic equation x2+6x+8 = 0 are
x=−2and x=−4.Question 4: Solve the following quadratic equation
for x:
x2+ 6x+ 8 = 0
Solution: To solve the quadratic equation x2+ 6x+ 8 = 0, we can
use the quadratic formula:
x=−b±√b2−4ac
2a
4
where a= 1,b= 6, and c= 8 in this case. Plug in these values into
the formula:
x=−6±√62−4·1·8
2·1
x=−6±√36 −32
2
x=−6±√4
2
Now, calculate the two possible solutions for xby splitting the ±:
x1=−6+2
2=−2
x2=−6−2
2=−4
Therefore, the solutions to the quadratic equation x2+ 6x+ 8 = 0
are x=−2and x=−4.
Question 5
Step-by-step solution: Step 1: Identify the coefficients of the
quadratic equation 2x2−5x−3=0. In this case, a= 2,b=−5, and
c=−3. Step 2: Use the quadratic formula x=−b±√b2−4ac
2ato solve
for x. Step 3: Substitute the values of a,b, and cinto the quadratic for-
mula: x=−(−5) ±p(−5)2−4·2·(−3)
2·2. Step 4: Simplify the equation:
x=5±√25 + 24
4. Step 5: Further simplify the equation: x=5±√49
4.
Step 6: Calculate the square root of 49: x=5±7
4. Step 7: Determine
the two possible solutions: - Solution 1: x=5+7
4=12
4= 3. - Solution
2: x=5−7
4=−2
4=−1
2. Therefore, the solutions to the quadratic
equation 2x2−5x−3=0are x= 3 and x=−1
2.Question 5: Solve the
following quadratic equation for x:2x2−5x−3=0.
Step-by-step solution: Step 1: Identify the coefficients of the
quadratic equation 2x2−5x−3=0. In this case, a= 2,b=−5, and
c=−3. Step 2: Use the quadratic formula x=−b±√b2−4ac
2ato solve
for x. Step 3: Substitute the values of a,b, and cinto the quadratic for-
mula: x=−(−5) ±p(−5)2−4·2·(−3)
2·2. Step 4: Simplify the equation:
5
x=5±√25 + 24
4. Step 5: Further simplify the equation: x=5±√49
4.
Step 6: Calculate the square root of 49: x=5±7
4. Step 7: Determine
the two possible solutions: - Solution 1: x=5+7
4=12
4= 3. - Solution
2: x=5−7
4=−2
4=−1
2. Therefore, the solutions to the quadratic
equation 2x2−5x−3=0are x= 3 and x=−1
2.
Question 6
Step-by-step solution: To solve the quadratic inequality 4x2−12x+
9≥0, we first find the roots of the associated quadratic equation
4x2−12x+ 9 = 0.
Step 1: Start by factoring the quadratic equation:
4x2−12x+ 9 = 0
(2x−3)2= 0
2x−3=0
2x= 3
x=3
2
Thus, the quadratic equation has one repeated root at x=3
2.
Step 2: Plot the critical point on the number line, x=3
2.
Step 3: Test a point in each of the intervals created by the critical
point (e.g., x= 0,x= 1) in the original inequality to determine the
sign of the expression: For x= 0:4(0)2−12(0) + 9 = 9 Since 9≥0, the
inequality holds for this interval.
For x= 1:4(1)2−12(1) + 9 = 1 Since 1≥0, the inequality holds for
this interval.
Step 4: Based on the tests, the solution to the quadratic inequality
4x2−12x+ 9 ≥0is x≤3
2.Question 6: Solve the following quadratic
inequality: 4x2−12x+ 9 ≥0
Step-by-step solution: To solve the quadratic inequality 4x2−12x+
9≥0, we first find the roots of the associated quadratic equation
4x2−12x+ 9 = 0.
Step 1: Start by factoring the quadratic equation:
4x2−12x+ 9 = 0
(2x−3)2= 0
2x−3=0
6
2x= 3
x=3
2
Thus, the quadratic equation has one repeated root at x=3
2.
Step 2: Plot the critical point on the number line, x=3
2.
Step 3: Test a point in each of the intervals created by the critical
point (e.g., x= 0,x= 1) in the original inequality to determine the
sign of the expression: For x= 0:4(0)2−12(0) + 9 = 9 Since 9≥0, the
inequality holds for this interval.
For x= 1:4(1)2−12(1) + 9 = 1 Since 1≥0, the inequality holds for
this interval.
Step 4: Based on the tests, the solution to the quadratic inequality
4x2−12x+ 9 ≥0is x≤3
2.
Question 7
Step-by-step solution: 1. First, factor the quadratic expression to
find the roots of the related equation 2x2−5x+ 2 = 0. The factored
form is (2x−1)(x−2) = 0, which gives the roots x=1
2and x= 2. 2.
Plot these roots on a number line to divide it into three intervals:
(−∞,1
2),(1
2,2), and (2,∞). 3. Choose a test point from each interval
to determine the sign of the inequality in that interval. For example,
for the interval (−∞,1
2), choose x= 0 as a test point. 4. Substitute
the test points back into the inequality 2x2−5x+ 2 >0and check the
sign of the expression to determine where it is positive. 5. Based on
the signs of these test points, we find that the inequality is satisfied
for x∈(1
2,2). 6. Therefore, the solution to the quadratic inequality
2x2−5x+ 2 >0is x∈(1
2,2).Question 7: Solve the quadratic inequality
2x2−5x+ 2 >0
Step-by-step solution: 1. First, factor the quadratic expression to
find the roots of the related equation 2x2−5x+ 2 = 0. The factored
form is (2x−1)(x−2) = 0, which gives the roots x=1
2and x= 2. 2.
Plot these roots on a number line to divide it into three intervals:
(−∞,1
2),(1
2,2), and (2,∞). 3. Choose a test point from each interval
to determine the sign of the inequality in that interval. For example,
for the interval (−∞,1
2), choose x= 0 as a test point. 4. Substitute
the test points back into the inequality 2x2−5x+ 2 >0and check the
sign of the expression to determine where it is positive. 5. Based on
the signs of these test points, we find that the inequality is satisfied
for x∈(1
2,2). 6. Therefore, the solution to the quadratic inequality
2x2−5x+ 2 >0is x∈(1
2,2).
7
Question 8
Step-by-step Solution: 1. Begin by finding the critical points of
the inequality by setting the quadratic expression equal to zero: 2x2−
3x−2=0. 2. To solve the quadratic equation 2x2−3x−2=0, use the
quadratic formula: x=−b±√b2−4ac
2a, where a= 2,b=−3, and c=−2.
3. Calculate the discriminant, b2−4ac for the quadratic formula:
(−3)2−4∗2∗(−2) = 9 + 16 = 25. 4. Since the discriminant is positive
(25 >0), the quadratic equation has two real roots. 5. Substitute
the values of a,b, and the discriminant into the quadratic formula
to find the two roots of the quadratic equation. 6. Once you have
found the roots, x1and x2, plot them on a number line to create
intervals. 7. Test each interval with a test point to determine the
sign of the quadratic expression in that interval. 8. The solution to
the inequality 2x2−3x−2<0will be the intervals where the sign of
the expression is negative. 9. Write the final solution as an interval
or union of intervals.Question 8: Solve the quadratic inequality 2x2−
3x−2<0.
Step-by-step Solution: 1. Begin by finding the critical points of
the inequality by setting the quadratic expression equal to zero: 2x2−
3x−2=0. 2. To solve the quadratic equation 2x2−3x−2=0, use the
quadratic formula: x=−b±√b2−4ac
2a, where a= 2,b=−3, and c=−2.
3. Calculate the discriminant, b2−4ac for the quadratic formula:
(−3)2−4∗2∗(−2) = 9 + 16 = 25. 4. Since the discriminant is positive
(25 >0), the quadratic equation has two real roots. 5. Substitute
the values of a,b, and the discriminant into the quadratic formula
to find the two roots of the quadratic equation. 6. Once you have
found the roots, x1and x2, plot them on a number line to create
intervals. 7. Test each interval with a test point to determine the
sign of the quadratic expression in that interval. 8. The solution to
the inequality 2x2−3x−2<0will be the intervals where the sign of
the expression is negative. 9. Write the final solution as an interval
or union of intervals.
Question 9
Solve the quadratic inequality: 2x2+ 5x−3<0.
Solution:
To solve the quadratic inequality 2x2+ 5x−3<0, we need to find
the values of xthat satisfy the inequality.
Step 1: Find the roots of the quadratic equation 2x2+ 5x−3=0
by factoring or using the quadratic formula.
The quadratic equation 2x2+ 5x−3=0can be factored as (2x−
1)(x+ 3) = 0.
8
Setting each factor to zero gives 2x−1 = 0 and x+ 3 = 0, which
leads to x=1
2and x=−3.
So, the roots of the quadratic equation are x=1
2and x=−3.
Step 2: Plot these roots on a number line to create intervals.
−3divides the number line into two intervals: (−∞,−3) and (−3,1
2).
Step 3: Test points within each interval to determine the sign of
the quadratic expression.
For x=−4(in (−∞,−3)):
Plugging x=−4into 2x2+5x−3gives 2(−4)2+5(−4)−3 = 32−20−3 =
9>0.
For x= 0 (in (−3,1
2)):
Plugging x= 0 into 2x2+ 5x−3gives 2(0)2+ 5(0) −3 = −3<0.
Step 4: Analyze the signs in the intervals based on the test points.
Since the inequality 2x2+ 5x−3<0is satisfied for x∈(−3,1
2), the
solution to the inequality is (−3,1
2).Question 9:
Solve the quadratic inequality: 2x2+ 5x−3<0.
Solution:
To solve the quadratic inequality 2x2+ 5x−3<0, we need to find
the values of xthat satisfy the inequality.
Step 1: Find the roots of the quadratic equation 2x2+ 5x−3=0
by factoring or using the quadratic formula.
The quadratic equation 2x2+ 5x−3=0can be factored as (2x−
1)(x+ 3) = 0.
Setting each factor to zero gives 2x−1 = 0 and x+ 3 = 0, which
leads to x=1
2and x=−3.
So, the roots of the quadratic equation are x=1
2and x=−3.
Step 2: Plot these roots on a number line to create intervals.
−3divides the number line into two intervals: (−∞,−3) and (−3,1
2).
Step 3: Test points within each interval to determine the sign of
the quadratic expression.
For x=−4(in (−∞,−3)):
Plugging x=−4into 2x2+5x−3gives 2(−4)2+5(−4)−3 = 32−20−3 =
9>0.
For x= 0 (in (−3,1
2)):
Plugging x= 0 into 2x2+ 5x−3gives 2(0)2+ 5(0) −3 = −3<0.
Step 4: Analyze the signs in the intervals based on the test points.
Since the inequality 2x2+ 5x−3<0is satisfied for x∈(−3,1
2), the
solution to the inequality is (−3,1
2).
Question 10
Step-by-step solution: 1. Add 5 to both sides of the inequality:
9
2x−5 + 5 >3 + 5 2. Simplify: 2x > 83. Divide by 2 on both sides to
isolate the variable: 2x
2>8
24. Solve for x:x > 4
Therefore, the solution to the inequality 2x−5>3is x > 4.Question
10: Solve the linear inequality: 2x−5>3
Step-by-step solution: 1. Add 5 to both sides of the inequality:
2x−5 + 5 >3 + 5 2. Simplify: 2x > 83. Divide by 2 on both sides to
isolate the variable: 2x
2>8
24. Solve for x:x > 4
Therefore, the solution to the inequality 2x−5>3is x > 4.
Question 11
Step-by-step solution: 1. Start by factoring the quadratic expres-
sion 3x2−4x > 0:
3x2−4x > 0
x(3x−4) >0
2. The critical points are when x= 0 and when 3x−4=0, which
gives x=4
3. 3. Plot these critical points on a number line and test
intervals to determine the solutions. 4. Test the interval (−,0) with
a test point x=−1:
(−1)(3(−1) −4) >0
−3+4>0
1>0
5. Test the interval (0,4
3)with a test point x= 1:
1(3(1) −4) >0
3−4<0
−1<0
6. Test the interval (4
3,)with a test point x= 2:
2(3(2) −4) >0
2(6 −4) >0
4>0
7. Therefore, the solution to the inequality 3x2−4x > 0is x < 0or
x > 4
3.Question 11: Solve the quadratic inequality 3x2−4x > 0.
Step-by-step solution: 1. Start by factoring the quadratic expres-
sion 3x2−4x > 0:
3x2−4x > 0
x(3x−4) >0
10
2. The critical points are when x= 0 and when 3x−4=0, which
gives x=4
3. 3. Plot these critical points on a number line and test
intervals to determine the solutions. 4. Test the interval (−,0) with
a test point x=−1:
(−1)(3(−1) −4) >0
−3+4>0
1>0
5. Test the interval (0,4
3)with a test point x= 1:
1(3(1) −4) >0
3−4<0
−1<0
6. Test the interval (4
3,)with a test point x= 2:
2(3(2) −4) >0
2(6 −4) >0
4>0
7. Therefore, the solution to the inequality 3x2−4x > 0is x < 0or
x > 4
3.
Question 12
Step-by-step solution: To solve the inequality 2x2+ 5x−3≤0, we
follow these steps:
1. Factor the quadratic expression: 2x2+ 5x−3 = (2x−1)(x+ 3).
2. Set each factor equal to zero and solve for x:
2x−1=0
2x= 1
x=1
2
and
x+ 3 = 0
x=−3
3. Mark the critical points x=−3and x=1
2on a number line.
4. Test each interval created by the critical points with a test value
(e.g., 0): - For x < −3: Test x=−4:(2(−4) −1)(−4 + 3) = (−9)(−1) >0.
11
- For −3<x< 1
2: Test x= 0:(2(0) −1)(0 + 3) = (−1)(3) <0. - For x > 1
2:
Test x= 1:(2(1) −1)(1 + 3) = (1)(4) >0.
5. Determine the solution: The inequality 2x2+ 5x−3≤0is true
for −∞ < x ≤ −3and 1
2≥x < ∞.
Therefore, the solution to the inequality is x∈(−∞,−3]∪1
2,∞.Question
12: Solve the quadratic inequality 2x2+ 5x−3≤0.
Step-by-step solution: To solve the inequality 2x2+ 5x−3≤0, we
follow these steps:
1. Factor the quadratic expression: 2x2+ 5x−3 = (2x−1)(x+ 3).
2. Set each factor equal to zero and solve for x:
2x−1=0
2x= 1
x=1
2
and
x+ 3 = 0
x=−3
3. Mark the critical points x=−3and x=1
2on a number line.
4. Test each interval created by the critical points with a test value
(e.g., 0): - For x < −3: Test x=−4:(2(−4) −1)(−4 + 3) = (−9)(−1) >0.
- For −3<x< 1
2: Test x= 0:(2(0) −1)(0 + 3) = (−1)(3) <0. - For x > 1
2:
Test x= 1:(2(1) −1)(1 + 3) = (1)(4) >0.
5. Determine the solution: The inequality 2x2+ 5x−3≤0is true
for −∞ < x ≤ −3and 1
2≥x < ∞.
Therefore, the solution to the inequality is x∈(−∞,−3] ∪1
2,∞.
Question 13
Solve the inequality:
2x−5<3(x+ 4)
Step-by-step Solution:
Given inequality:
2x−5<3(x+ 4)
Expand the right side:
2x−5<3x+ 12
Subtract 2xfrom both sides:
12
−5< x + 12
Subtract 12 from both sides:
−17 < x
Therefore, the solution to the inequality is x > −17.Question 13:
Solve the inequality:
2x−5<3(x+ 4)
Step-by-step Solution:
Given inequality:
2x−5<3(x+ 4)
Expand the right side:
2x−5<3x+ 12
Subtract 2xfrom both sides:
−5< x + 12
Subtract 12 from both sides:
−17 < x
Therefore, the solution to the inequality is x > −17.
Question 14
Step-by-step solution: 1. Rewrite the inequality in standard form:
2x2−5x−3<0. 2. Factor the quadratic expression: (2x+ 1)(x−3) <0.
3. Determine the critical points by setting each factor equal to zero:
2x+ 1 = 0 and x−3=0. 4. Solve for x:2x=−1⇒x=−1
2and x= 3.
5. Plot the critical points on a number line and test intervals created
by these points in the original inequality. 6. Test interval 1: Choose
a test point x= 0. Substitute into the expression: 2(0)2−5(0) −3<0,
which simplifies to −3<0, and is true. Therefore, interval 1 is part of
the solution. 7. Test interval 2: Choose a test point x= 4. Substitute
into the expression: 2(4)2−5(4) −3<0, which simplifies to 17 <0,
and is false. Therefore, interval 2 is not part of the solution. 8. The
solution to the quadratic inequality is −1
2< x < 3.Question 14: Solve
the quadratic inequality 2x2−5x < 3.
Step-by-step solution: 1. Rewrite the inequality in standard form:
2x2−5x−3<0. 2. Factor the quadratic expression: (2x+ 1)(x−3) <0.
3. Determine the critical points by setting each factor equal to zero:
13
2x+ 1 = 0 and x−3=0. 4. Solve for x:2x=−1⇒x=−1
2and x= 3.
5. Plot the critical points on a number line and test intervals created
by these points in the original inequality. 6. Test interval 1: Choose
a test point x= 0. Substitute into the expression: 2(0)2−5(0) −3<0,
which simplifies to −3<0, and is true. Therefore, interval 1 is part of
the solution. 7. Test interval 2: Choose a test point x= 4. Substitute
into the expression: 2(4)2−5(4) −3<0, which simplifies to 17 <0,
and is false. Therefore, interval 2 is not part of the solution. 8. The
solution to the quadratic inequality is −1
2<x<3.
Question 15
3x2−5x > 2
Step-by-step solution:
1. Start by moving all terms to one side of the inequality to set it
equal to zero:
3x2−5x−2>0
2. Next, factor the quadratic expression:
3x2−5x−2 = (3x+ 1)(x−2) >0
3. Find the critical points by setting each factor equal to zero:
3x+ 1 = 0 →x=−1
3
x−2=0→x= 2
4. Create a number line and plot the critical points on it:
−1
32
−0 +
5. Test each interval on the number line by choosing a test point
in each interval and checking its sign when substituted into the in-
equality:
For x < −1
3: Choose x=−1
3(−1)2−5(−1) −2=8>0
So, this interval is positive.
For −1
3<x<2: Choose x= 0
3(0)2−5(0) −2 = −2<0
So, this interval is negative.
For x > 2: Choose x= 3
3(3)2−5(3) −2 = 16 >0
So, this interval is positive.
6. Determine the solution by analyzing the signs on the number
line:
The solution to the inequality 3x2−5x−2>0is −1
3<x<2, which
can be written in interval notation as (−1
3,2).Question 15: Solve the
following quadratic inequality:
3x2−5x > 2
Step-by-step solution:
14
1. Start by moving all terms to one side of the inequality to set it
equal to zero:
3x2−5x−2>0
2. Next, factor the quadratic expression:
3x2−5x−2 = (3x+ 1)(x−2) >0
3. Find the critical points by setting each factor equal to zero:
3x+ 1 = 0 →x=−1
3
x−2=0→x= 2
4. Create a number line and plot the critical points on it:
−1
32
−0 +
5. Test each interval on the number line by choosing a test point
in each interval and checking its sign when substituted into the in-
equality:
For x < −1
3: Choose x=−1
3(−1)2−5(−1) −2=8>0
So, this interval is positive.
For −1
3<x<2: Choose x= 0
3(0)2−5(0) −2 = −2<0
So, this interval is negative.
For x > 2: Choose x= 3
3(3)2−5(3) −2 = 16 >0
So, this interval is positive.
6. Determine the solution by analyzing the signs on the number
line:
The solution to the inequality 3x2−5x−2>0is −1
3<x<2, which
can be written in interval notation as (−1
3,2).
Question 16
Step-by-step solution: 1. Start by finding the roots of the corre-
sponding quadratic equation 2x2−7x+ 3 = 0. 2. Use the quadratic
formula x=−b±√b2−4ac
2awhere a= 2,b=−7, and c= 3 to find the roots.
3. Calculate the discriminant ∆ = b2−4ac to determine the nature
of the roots. 4. If the discriminant is positive, there are two distinct
real roots. If it is zero, there is one real root. If it is negative, there
are no real roots. 5. Once you have found the roots, plot them on a
number line to divide the number line into intervals. 6. Test a value
from each interval in the original inequality to determine the solution
set. 7. The solution to the quadratic inequality 2x2−7x+ 3 <0will
be the values that make the inequality true.Question 16: Solve the
quadratic inequality 2x2−7x+ 3 <0.
Step-by-step solution: 1. Start by finding the roots of the corre-
sponding quadratic equation 2x2−7x+ 3 = 0. 2. Use the quadratic
15
formula x=−b±√b2−4ac
2awhere a= 2,b=−7, and c= 3 to find the roots.
3. Calculate the discriminant ∆ = b2−4ac to determine the nature
of the roots. 4. If the discriminant is positive, there are two distinct
real roots. If it is zero, there is one real root. If it is negative, there
are no real roots. 5. Once you have found the roots, plot them on a
number line to divide the number line into intervals. 6. Test a value
from each interval in the original inequality to determine the solution
set. 7. The solution to the quadratic inequality 2x2−7x+ 3 <0will
be the values that make the inequality true.
Question 17
Step-by-step solution:
3
2x−1≤1
2(4x+ 7)
3x−2≤2x+7
2(Distribute on the right side)
3x−2x≤7
2+ 2 (Subtract 2xfrom both sides)
x≤11
2
Therefore, the solution to the inequality is x≤11
2.Question 17:
Solve the inequality 3
2x−1≤1
2(4x+ 7).
Step-by-step solution:
3
2x−1≤1
2(4x+ 7)
3x−2≤2x+7
2(Distribute on the right side)
3x−2x≤7
2+ 2 (Subtract 2xfrom both sides)
x≤11
2
Therefore, the solution to the inequality is x≤11
2.
Question 18
Solve the following quadratic inequality:
(x−3)(x+ 2) ≥0
Solution:
16
To solve the quadratic inequality (x−3)(x+ 2) ≥0, we need to find
the critical points where the expression equals zero and determine
the sign of the expression in each interval.
1. Find the critical points by setting the expression equal to zero:
(x−3)(x+ 2) = 0
This gives us x= 3 and x=−2as critical points.
2. Plot these critical points on a number line:
−2 3
3. Test the intervals created by the critical points by choosing test
points and determining the sign of the expression in each interval.
- For x < −2, choose x=−3:
(−3−3)(−3 + 2) = (−6)(−1) = 6 >0
Since the expression is positive, solutions in this interval are included.
- For −2<x<3, choose x= 0:
(0 −3)(0 + 2) = (−3)(2) = −6<0
Since the expression is negative, solutions in this interval are not
included.
- For x > 3, choose x= 4:
(4 −3)(4 + 2) = (1)(6) = 6 >0
Since the expression is positive, solutions in this interval are included.
4. Combine the results to write the solution set:
x∈(−∞,−2] ∪[3,∞)
Therefore, the solution to the inequality (x−3)(x+ 2) ≥0is x∈
(−∞,−2] ∪[3,∞).Question 18:
Solve the following quadratic inequality:
(x−3)(x+ 2) ≥0
Solution:
To solve the quadratic inequality (x−3)(x+ 2) ≥0, we need to find
the critical points where the expression equals zero and determine
the sign of the expression in each interval.
1. Find the critical points by setting the expression equal to zero:
(x−3)(x+ 2) = 0
This gives us x= 3 and x=−2as critical points.
17
2. Plot these critical points on a number line:
−2 3
3. Test the intervals created by the critical points by choosing test
points and determining the sign of the expression in each interval.
- For x < −2, choose x=−3:
(−3−3)(−3 + 2) = (−6)(−1) = 6 >0
Since the expression is positive, solutions in this interval are included.
- For −2<x<3, choose x= 0:
(0 −3)(0 + 2) = (−3)(2) = −6<0
Since the expression is negative, solutions in this interval are not
included.
- For x > 3, choose x= 4:
(4 −3)(4 + 2) = (1)(6) = 6 >0
Since the expression is positive, solutions in this interval are included.
4. Combine the results to write the solution set:
x∈(−∞,−2] ∪[3,∞)
Therefore, the solution to the inequality (x−3)(x+ 2) ≥0is x∈
(−∞,−2] ∪[3,∞).
Question 19
Solve the following quadratic inequality:
2x2−5x+ 2 <0
Step-by-Step Solution:
To solve the given quadratic inequality, we first rewrite it in the
standard form:
2x2−5x+ 2 <0
Next, we factor the quadratic expression:
2x2−4x−x+ 2 <0
2x(x−2) −1(x−2) <0
(2x−1)(x−2) <0
18
Now, we find the critical points by setting each factor equal to
zero:
2x−1=0⇒x=1
2
x−2 = 0 ⇒x= 2
These critical points divide the real number line into three inter-
vals: (−∞,1
2),(1
2,2),(2,+∞).
We test each interval by choosing a test point and plugging it into
the inequality to determine the sign of the expression:
For x= 0:(2(0) −1)(0 −2) = (−1)(−2) = 2 >0, which is false.
For x= 1:(2(1) −1)(1 −2) = (1)(−1) = −1<0, which is true.
For x= 3:(2(3) −1)(3 −2) = (5)(1) = 5 >0, which is false.
Therefore, the solution to the inequality 2x2−5x+ 2 <0is:
x∈1
2,2
Question 19:
Solve the following quadratic inequality:
2x2−5x+ 2 <0
Step-by-Step Solution:
To solve the given quadratic inequality, we first rewrite it in the
standard form:
2x2−5x+ 2 <0
Next, we factor the quadratic expression:
2x2−4x−x+ 2 <0
2x(x−2) −1(x−2) <0
(2x−1)(x−2) <0
Now, we find the critical points by setting each factor equal to
zero:
2x−1=0⇒x=1
2
x−2 = 0 ⇒x= 2
These critical points divide the real number line into three inter-
vals: (−∞,1
2),(1
2,2),(2,+∞).
We test each interval by choosing a test point and plugging it into
the inequality to determine the sign of the expression:
For x= 0:(2(0) −1)(0 −2) = (−1)(−2) = 2 >0, which is false.
For x= 1:(2(1) −1)(1 −2) = (1)(−1) = −1<0, which is true.
For x= 3:(2(3) −1)(3 −2) = (5)(1) = 5 >0, which is false.
19
Therefore, the solution to the inequality 2x2−5x+ 2 <0is:
x∈1
2,2
Question 20
3x2+ 4x−4>0
Step-by-step solution: 1. Begin by rewriting the inequality in
standard form:
3x2+ 4x−4>0
2. Identify the coefficients of the quadratic equation:
a= 3, b = 4, c =−4
3. Calculate the discriminant:
∆ = b2−4ac
∆=42−4(3)(−4)
∆ = 16 + 48
∆ = 64
4. Determine the nature of the roots using the discriminant: - If
∆>0, the inequality will have two distinct real roots. - If ∆=0, the
inequality will have a repeated real root. - If ∆<0, the inequality
will have no real roots.
5. Since ∆>0, the inequality will have two distinct real roots. To
find the roots, use the quadratic formula:
x=−b±√∆
2a
x=−4±√64
2(3)
x=−4±8
6
6. We get two roots:
x1=−4+8
6=4
6=2
3
x2=−4−8
6=−12
6=−2
20
7. Now, we need to determine the intervals where the inequality is
satisfied. To do this, we create a sign chart using the roots we found.
x < −2−2<x< 2
3x > 2
3
3x2+ 4x−4−+ +
8. Since the inequality is greater than zero (>0), the solution is
where the quadratic expression is positive:
x∈(−∞,−2) ∪2
3,∞
Therefore, the solution set for the quadratic inequality 3x2+ 4x−
4>0is (−∞,−2) ∪2
3,∞.Question 20: Solve the following quadratic
inequality:
3x2+ 4x−4>0
Step-by-step solution: 1. Begin by rewriting the inequality in
standard form:
3x2+ 4x−4>0
2. Identify the coefficients of the quadratic equation:
a= 3, b = 4, c =−4
3. Calculate the discriminant:
∆ = b2−4ac
∆=42−4(3)(−4)
∆ = 16 + 48
∆ = 64
4. Determine the nature of the roots using the discriminant: - If
∆>0, the inequality will have two distinct real roots. - If ∆=0, the
inequality will have a repeated real root. - If ∆<0, the inequality
will have no real roots.
5. Since ∆>0, the inequality will have two distinct real roots. To
find the roots, use the quadratic formula:
x=−b±√∆
2a
x=−4±√64
2(3)
x=−4±8
6
21
Question 2
Solve the following linear equation for x:
3(2x−4) = 5x+ 7
Solution:
Step 1: Distribute the 3 on the left side of the equation.
3(2x−4) = 5x+ 7
6x−12 = 5x+ 7
Step 2: Move all terms involving xto one side of the equation and
constants to the other side.
6x−12 = 5x+ 7
6x−5x= 7 + 12
x= 19
Therefore, the solution to the equation 3(2x−4) = 5x+ 7 is x=
19.Question 2:
Solve the following linear equation for x:
3(2x−4) = 5x+ 7
Solution:
Step 1: Distribute the 3 on the left side of the equation.
3(2x−4) = 5x+ 7
6x−12 = 5x+ 7
Step 2: Move all terms involving xto one side of the equation and
constants to the other side.
6x−12 = 5x+ 7
6x−5x= 7 + 12
x= 19
Therefore, the solution to the equation 3(2x−4) = 5x+ 7 is x= 19.
Question 3
Solution: To solve the quadratic equation 2x2+ 5x−3=0, we can
use the quadratic formula: The quadratic formula is given by:
x=−b±√b2−4ac
2a
2
Given the quadratic equation 2x2+ 5x−3=0, we have: a= 2,b= 5,
and c=−3
Plugging these values into the quadratic formula, we get:
x=−5±p52−4(2)(−3)
2(2)
x=−5±√25 + 24
4
x=−5±√49
4
x=−5±7
4
Therefore, the solutions to the quadratic equation 2x2+ 5x−3=0
are:
x=−5+7
4=2
4=1
2
and
x=−5−7
4=−12
4=−3
So, the solutions to the quadratic equation 2x2+ 5x−3=0are x=
1
2and x=−3.Question 3: Solve the following quadratic equation:
2x2+ 5x−3=0
Solution: To solve the quadratic equation 2x2+ 5x−3=0, we can
use the quadratic formula: The quadratic formula is given by:
x=−b±√b2−4ac
2a
Given the quadratic equation 2x2+ 5x−3=0, we have: a= 2,b= 5,
and c=−3
Plugging these values into the quadratic formula, we get:
x=−5±p52−4(2)(−3)
2(2)
x=−5±√25 + 24
4
x=−5±√49
4
x=−5±7
4
3
Therefore, the solutions to the quadratic equation 2x2+ 5x−3=0
are:
x=−5+7
4=2
4=1
2
and
x=−5−7
4=−12
4=−3
So, the solutions to the quadratic equation 2x2+ 5x−3 = 0 are x=1
2
and x=−3.
Question 4
Solution: To solve the quadratic equation x2+ 6x+ 8 = 0, we can
use the quadratic formula:
x=−b±√b2−4ac
2a
where a= 1,b= 6, and c= 8 in this case. Plug in these values into
the formula:
x=−6±√62−4·1·8
2·1
x=−6±√36 −32
2
x=−6±√4
2
Now, calculate the two possible solutions for xby splitting the ±:
x1=−6+2
2=−2
x2=−6−2
2=−4
Therefore, the solutions to the quadratic equation x2+6x+8 = 0 are
x=−2and x=−4.Question 4: Solve the following quadratic equation
for x:
x2+ 6x+ 8 = 0
Solution: To solve the quadratic equation x2+ 6x+ 8 = 0, we can
use the quadratic formula:
x=−b±√b2−4ac
2a
4
where a= 1,b= 6, and c= 8 in this case. Plug in these values into
the formula:
x=−6±√62−4·1·8
2·1
x=−6±√36 −32
2
x=−6±√4
2
Now, calculate the two possible solutions for xby splitting the ±:
x1=−6+2
2=−2
x2=−6−2
2=−4
Therefore, the solutions to the quadratic equation x2+ 6x+ 8 = 0
are x=−2and x=−4.
Question 5
Step-by-step solution: Step 1: Identify the coefficients of the
quadratic equation 2x2−5x−3=0. In this case, a= 2,b=−5, and
c=−3. Step 2: Use the quadratic formula x=−b±√b2−4ac
2ato solve
for x. Step 3: Substitute the values of a,b, and cinto the quadratic for-
mula: x=−(−5) ±p(−5)2−4·2·(−3)
2·2. Step 4: Simplify the equation:
x=5±√25 + 24
4. Step 5: Further simplify the equation: x=5±√49
4.
Step 6: Calculate the square root of 49: x=5±7
4. Step 7: Determine
the two possible solutions: - Solution 1: x=5+7
4=12
4= 3. - Solution
2: x=5−7
4=−2
4=−1
2. Therefore, the solutions to the quadratic
equation 2x2−5x−3=0are x= 3 and x=−1
2.Question 5: Solve the
following quadratic equation for x:2x2−5x−3=0.
Step-by-step solution: Step 1: Identify the coefficients of the
quadratic equation 2x2−5x−3=0. In this case, a= 2,b=−5, and
c=−3. Step 2: Use the quadratic formula x=−b±√b2−4ac
2ato solve
for x. Step 3: Substitute the values of a,b, and cinto the quadratic for-
mula: x=−(−5) ±p(−5)2−4·2·(−3)
2·2. Step 4: Simplify the equation:
5
x=5±√25 + 24
4. Step 5: Further simplify the equation: x=5±√49
4.
Step 6: Calculate the square root of 49: x=5±7
4. Step 7: Determine
the two possible solutions: - Solution 1: x=5+7
4=12
4= 3. - Solution
2: x=5−7
4=−2
4=−1
2. Therefore, the solutions to the quadratic
equation 2x2−5x−3=0are x= 3 and x=−1
2.
Question 6
Step-by-step solution: To solve the quadratic inequality 4x2−12x+
9≥0, we first find the roots of the associated quadratic equation
4x2−12x+ 9 = 0.
Step 1: Start by factoring the quadratic equation:
4x2−12x+ 9 = 0
(2x−3)2= 0
2x−3=0
2x= 3
x=3
2
Thus, the quadratic equation has one repeated root at x=3
2.
Step 2: Plot the critical point on the number line, x=3
2.
Step 3: Test a point in each of the intervals created by the critical
point (e.g., x= 0,x= 1) in the original inequality to determine the
sign of the expression: For x= 0:4(0)2−12(0) + 9 = 9 Since 9≥0, the
inequality holds for this interval.
For x= 1:4(1)2−12(1) + 9 = 1 Since 1≥0, the inequality holds for
this interval.
Step 4: Based on the tests, the solution to the quadratic inequality
4x2−12x+ 9 ≥0is x≤3
2.Question 6: Solve the following quadratic
inequality: 4x2−12x+ 9 ≥0
Step-by-step solution: To solve the quadratic inequality 4x2−12x+
9≥0, we first find the roots of the associated quadratic equation
4x2−12x+ 9 = 0.
Step 1: Start by factoring the quadratic equation:
4x2−12x+ 9 = 0
(2x−3)2= 0
2x−3=0
6
2x= 3
x=3
2
Thus, the quadratic equation has one repeated root at x=3
2.
Step 2: Plot the critical point on the number line, x=3
2.
Step 3: Test a point in each of the intervals created by the critical
point (e.g., x= 0,x= 1) in the original inequality to determine the
sign of the expression: For x= 0:4(0)2−12(0) + 9 = 9 Since 9≥0, the
inequality holds for this interval.
For x= 1:4(1)2−12(1) + 9 = 1 Since 1≥0, the inequality holds for
this interval.
Step 4: Based on the tests, the solution to the quadratic inequality
4x2−12x+ 9 ≥0is x≤3
2.
Question 7
Step-by-step solution: 1. First, factor the quadratic expression to
find the roots of the related equation 2x2−5x+ 2 = 0. The factored
form is (2x−1)(x−2) = 0, which gives the roots x=1
2and x= 2. 2.
Plot these roots on a number line to divide it into three intervals:
(−∞,1
2),(1
2,2), and (2,∞). 3. Choose a test point from each interval
to determine the sign of the inequality in that interval. For example,
for the interval (−∞,1
2), choose x= 0 as a test point. 4. Substitute
the test points back into the inequality 2x2−5x+ 2 >0and check the
sign of the expression to determine where it is positive. 5. Based on
the signs of these test points, we find that the inequality is satisfied
for x∈(1
2,2). 6. Therefore, the solution to the quadratic inequality
2x2−5x+ 2 >0is x∈(1
2,2).Question 7: Solve the quadratic inequality
2x2−5x+ 2 >0
Step-by-step solution: 1. First, factor the quadratic expression to
find the roots of the related equation 2x2−5x+ 2 = 0. The factored
form is (2x−1)(x−2) = 0, which gives the roots x=1
2and x= 2. 2.
Plot these roots on a number line to divide it into three intervals:
(−∞,1
2),(1
2,2), and (2,∞). 3. Choose a test point from each interval
to determine the sign of the inequality in that interval. For example,
for the interval (−∞,1
2), choose x= 0 as a test point. 4. Substitute
the test points back into the inequality 2x2−5x+ 2 >0and check the
sign of the expression to determine where it is positive. 5. Based on
the signs of these test points, we find that the inequality is satisfied
for x∈(1
2,2). 6. Therefore, the solution to the quadratic inequality
2x2−5x+ 2 >0is x∈(1
2,2).
7
Question 8
Step-by-step Solution: 1. Begin by finding the critical points of
the inequality by setting the quadratic expression equal to zero: 2x2−
3x−2=0. 2. To solve the quadratic equation 2x2−3x−2=0, use the
quadratic formula: x=−b±√b2−4ac
2a, where a= 2,b=−3, and c=−2.
3. Calculate the discriminant, b2−4ac for the quadratic formula:
(−3)2−4∗2∗(−2) = 9 + 16 = 25. 4. Since the discriminant is positive
(25 >0), the quadratic equation has two real roots. 5. Substitute
the values of a,b, and the discriminant into the quadratic formula
to find the two roots of the quadratic equation. 6. Once you have
found the roots, x1and x2, plot them on a number line to create
intervals. 7. Test each interval with a test point to determine the
sign of the quadratic expression in that interval. 8. The solution to
the inequality 2x2−3x−2<0will be the intervals where the sign of
the expression is negative. 9. Write the final solution as an interval
or union of intervals.Question 8: Solve the quadratic inequality 2x2−
3x−2<0.
Step-by-step Solution: 1. Begin by finding the critical points of
the inequality by setting the quadratic expression equal to zero: 2x2−
3x−2=0. 2. To solve the quadratic equation 2x2−3x−2=0, use the
quadratic formula: x=−b±√b2−4ac
2a, where a= 2,b=−3, and c=−2.
3. Calculate the discriminant, b2−4ac for the quadratic formula:
(−3)2−4∗2∗(−2) = 9 + 16 = 25. 4. Since the discriminant is positive
(25 >0), the quadratic equation has two real roots. 5. Substitute
the values of a,b, and the discriminant into the quadratic formula
to find the two roots of the quadratic equation. 6. Once you have
found the roots, x1and x2, plot them on a number line to create
intervals. 7. Test each interval with a test point to determine the
sign of the quadratic expression in that interval. 8. The solution to
the inequality 2x2−3x−2<0will be the intervals where the sign of
the expression is negative. 9. Write the final solution as an interval
or union of intervals.
Question 9
Solve the quadratic inequality: 2x2+ 5x−3<0.
Solution:
To solve the quadratic inequality 2x2+ 5x−3<0, we need to find
the values of xthat satisfy the inequality.
Step 1: Find the roots of the quadratic equation 2x2+ 5x−3=0
by factoring or using the quadratic formula.
The quadratic equation 2x2+ 5x−3=0can be factored as (2x−
1)(x+ 3) = 0.
8
Setting each factor to zero gives 2x−1 = 0 and x+ 3 = 0, which
leads to x=1
2and x=−3.
So, the roots of the quadratic equation are x=1
2and x=−3.
Step 2: Plot these roots on a number line to create intervals.
−3divides the number line into two intervals: (−∞,−3) and (−3,1
2).
Step 3: Test points within each interval to determine the sign of
the quadratic expression.
For x=−4(in (−∞,−3)):
Plugging x=−4into 2x2+5x−3gives 2(−4)2+5(−4)−3 = 32−20−3 =
9>0.
For x= 0 (in (−3,1
2)):
Plugging x= 0 into 2x2+ 5x−3gives 2(0)2+ 5(0) −3 = −3<0.
Step 4: Analyze the signs in the intervals based on the test points.
Since the inequality 2x2+ 5x−3<0is satisfied for x∈(−3,1
2), the
solution to the inequality is (−3,1
2).Question 9:
Solve the quadratic inequality: 2x2+ 5x−3<0.
Solution:
To solve the quadratic inequality 2x2+ 5x−3<0, we need to find
the values of xthat satisfy the inequality.
Step 1: Find the roots of the quadratic equation 2x2+ 5x−3=0
by factoring or using the quadratic formula.
The quadratic equation 2x2+ 5x−3=0can be factored as (2x−
1)(x+ 3) = 0.
Setting each factor to zero gives 2x−1 = 0 and x+ 3 = 0, which
leads to x=1
2and x=−3.
So, the roots of the quadratic equation are x=1
2and x=−3.
Step 2: Plot these roots on a number line to create intervals.
−3divides the number line into two intervals: (−∞,−3) and (−3,1
2).
Step 3: Test points within each interval to determine the sign of
the quadratic expression.
For x=−4(in (−∞,−3)):
Plugging x=−4into 2x2+5x−3gives 2(−4)2+5(−4)−3 = 32−20−3 =
9>0.
For x= 0 (in (−3,1
2)):
Plugging x= 0 into 2x2+ 5x−3gives 2(0)2+ 5(0) −3 = −3<0.
Step 4: Analyze the signs in the intervals based on the test points.
Since the inequality 2x2+ 5x−3<0is satisfied for x∈(−3,1
2), the
solution to the inequality is (−3,1
2).
Question 10
Step-by-step solution: 1. Add 5 to both sides of the inequality:
9
2x−5 + 5 >3 + 5 2. Simplify: 2x > 83. Divide by 2 on both sides to
isolate the variable: 2x
2>8
24. Solve for x:x > 4
Therefore, the solution to the inequality 2x−5>3is x > 4.Question
10: Solve the linear inequality: 2x−5>3
Step-by-step solution: 1. Add 5 to both sides of the inequality:
2x−5 + 5 >3 + 5 2. Simplify: 2x > 83. Divide by 2 on both sides to
isolate the variable: 2x
2>8
24. Solve for x:x > 4
Therefore, the solution to the inequality 2x−5>3is x > 4.
Question 11
Step-by-step solution: 1. Start by factoring the quadratic expres-
sion 3x2−4x > 0:
3x2−4x > 0
x(3x−4) >0
2. The critical points are when x= 0 and when 3x−4=0, which
gives x=4
3. 3. Plot these critical points on a number line and test
intervals to determine the solutions. 4. Test the interval (−,0) with
a test point x=−1:
(−1)(3(−1) −4) >0
−3+4>0
1>0
5. Test the interval (0,4
3)with a test point x= 1:
1(3(1) −4) >0
3−4<0
−1<0
6. Test the interval (4
3,)with a test point x= 2:
2(3(2) −4) >0
2(6 −4) >0
4>0
7. Therefore, the solution to the inequality 3x2−4x > 0is x < 0or
x > 4
3.Question 11: Solve the quadratic inequality 3x2−4x > 0.
Step-by-step solution: 1. Start by factoring the quadratic expres-
sion 3x2−4x > 0:
3x2−4x > 0
x(3x−4) >0
10
2. The critical points are when x= 0 and when 3x−4=0, which
gives x=4
3. 3. Plot these critical points on a number line and test
intervals to determine the solutions. 4. Test the interval (−,0) with
a test point x=−1:
(−1)(3(−1) −4) >0
−3+4>0
1>0
5. Test the interval (0,4
3)with a test point x= 1:
1(3(1) −4) >0
3−4<0
−1<0
6. Test the interval (4
3,)with a test point x= 2:
2(3(2) −4) >0
2(6 −4) >0
4>0
7. Therefore, the solution to the inequality 3x2−4x > 0is x < 0or
x > 4
3.
Question 12
Step-by-step solution: To solve the inequality 2x2+ 5x−3≤0, we
follow these steps:
1. Factor the quadratic expression: 2x2+ 5x−3 = (2x−1)(x+ 3).
2. Set each factor equal to zero and solve for x:
2x−1=0
2x= 1
x=1
2
and
x+ 3 = 0
x=−3
3. Mark the critical points x=−3and x=1
2on a number line.
4. Test each interval created by the critical points with a test value
(e.g., 0): - For x < −3: Test x=−4:(2(−4) −1)(−4 + 3) = (−9)(−1) >0.
11
- For −3<x< 1
2: Test x= 0:(2(0) −1)(0 + 3) = (−1)(3) <0. - For x > 1
2:
Test x= 1:(2(1) −1)(1 + 3) = (1)(4) >0.
5. Determine the solution: The inequality 2x2+ 5x−3≤0is true
for −∞ < x ≤ −3and 1
2≥x < ∞.
Therefore, the solution to the inequality is x∈(−∞,−3]∪1
2,∞.Question
12: Solve the quadratic inequality 2x2+ 5x−3≤0.
Step-by-step solution: To solve the inequality 2x2+ 5x−3≤0, we
follow these steps:
1. Factor the quadratic expression: 2x2+ 5x−3 = (2x−1)(x+ 3).
2. Set each factor equal to zero and solve for x:
2x−1=0
2x= 1
x=1
2
and
x+ 3 = 0
x=−3
3. Mark the critical points x=−3and x=1
2on a number line.
4. Test each interval created by the critical points with a test value
(e.g., 0): - For x < −3: Test x=−4:(2(−4) −1)(−4 + 3) = (−9)(−1) >0.
- For −3<x< 1
2: Test x= 0:(2(0) −1)(0 + 3) = (−1)(3) <0. - For x > 1
2:
Test x= 1:(2(1) −1)(1 + 3) = (1)(4) >0.
5. Determine the solution: The inequality 2x2+ 5x−3≤0is true
for −∞ < x ≤ −3and 1
2≥x < ∞.
Therefore, the solution to the inequality is x∈(−∞,−3] ∪1
2,∞.
Question 13
Solve the inequality:
2x−5<3(x+ 4)
Step-by-step Solution:
Given inequality:
2x−5<3(x+ 4)
Expand the right side:
2x−5<3x+ 12
Subtract 2xfrom both sides:
12
−5< x + 12
Subtract 12 from both sides:
−17 < x
Therefore, the solution to the inequality is x > −17.Question 13:
Solve the inequality:
2x−5<3(x+ 4)
Step-by-step Solution:
Given inequality:
2x−5<3(x+ 4)
Expand the right side:
2x−5<3x+ 12
Subtract 2xfrom both sides:
−5< x + 12
Subtract 12 from both sides:
−17 < x
Therefore, the solution to the inequality is x > −17.
Question 14
Step-by-step solution: 1. Rewrite the inequality in standard form:
2x2−5x−3<0. 2. Factor the quadratic expression: (2x+ 1)(x−3) <0.
3. Determine the critical points by setting each factor equal to zero:
2x+ 1 = 0 and x−3=0. 4. Solve for x:2x=−1⇒x=−1
2and x= 3.
5. Plot the critical points on a number line and test intervals created
by these points in the original inequality. 6. Test interval 1: Choose
a test point x= 0. Substitute into the expression: 2(0)2−5(0) −3<0,
which simplifies to −3<0, and is true. Therefore, interval 1 is part of
the solution. 7. Test interval 2: Choose a test point x= 4. Substitute
into the expression: 2(4)2−5(4) −3<0, which simplifies to 17 <0,
and is false. Therefore, interval 2 is not part of the solution. 8. The
solution to the quadratic inequality is −1
2< x < 3.Question 14: Solve
the quadratic inequality 2x2−5x < 3.
Step-by-step solution: 1. Rewrite the inequality in standard form:
2x2−5x−3<0. 2. Factor the quadratic expression: (2x+ 1)(x−3) <0.
3. Determine the critical points by setting each factor equal to zero:
13
2x+ 1 = 0 and x−3=0. 4. Solve for x:2x=−1⇒x=−1
2and x= 3.
5. Plot the critical points on a number line and test intervals created
by these points in the original inequality. 6. Test interval 1: Choose
a test point x= 0. Substitute into the expression: 2(0)2−5(0) −3<0,
which simplifies to −3<0, and is true. Therefore, interval 1 is part of
the solution. 7. Test interval 2: Choose a test point x= 4. Substitute
into the expression: 2(4)2−5(4) −3<0, which simplifies to 17 <0,
and is false. Therefore, interval 2 is not part of the solution. 8. The
solution to the quadratic inequality is −1
2<x<3.
Question 15
3x2−5x > 2
Step-by-step solution:
1. Start by moving all terms to one side of the inequality to set it
equal to zero:
3x2−5x−2>0
2. Next, factor the quadratic expression:
3x2−5x−2 = (3x+ 1)(x−2) >0
3. Find the critical points by setting each factor equal to zero:
3x+ 1 = 0 →x=−1
3
x−2=0→x= 2
4. Create a number line and plot the critical points on it:
−1
32
−0 +
5. Test each interval on the number line by choosing a test point
in each interval and checking its sign when substituted into the in-
equality:
For x < −1
3: Choose x=−1
3(−1)2−5(−1) −2=8>0
So, this interval is positive.
For −1
3<x<2: Choose x= 0
3(0)2−5(0) −2 = −2<0
So, this interval is negative.
For x > 2: Choose x= 3
3(3)2−5(3) −2 = 16 >0
So, this interval is positive.
6. Determine the solution by analyzing the signs on the number
line:
The solution to the inequality 3x2−5x−2>0is −1
3<x<2, which
can be written in interval notation as (−1
3,2).Question 15: Solve the
following quadratic inequality:
3x2−5x > 2
Step-by-step solution:
14
1. Start by moving all terms to one side of the inequality to set it
equal to zero:
3x2−5x−2>0
2. Next, factor the quadratic expression:
3x2−5x−2 = (3x+ 1)(x−2) >0
3. Find the critical points by setting each factor equal to zero:
3x+ 1 = 0 →x=−1
3
x−2=0→x= 2
4. Create a number line and plot the critical points on it:
−1
32
−0 +
5. Test each interval on the number line by choosing a test point
in each interval and checking its sign when substituted into the in-
equality:
For x < −1
3: Choose x=−1
3(−1)2−5(−1) −2=8>0
So, this interval is positive.
For −1
3<x<2: Choose x= 0
3(0)2−5(0) −2 = −2<0
So, this interval is negative.
For x > 2: Choose x= 3
3(3)2−5(3) −2 = 16 >0
So, this interval is positive.
6. Determine the solution by analyzing the signs on the number
line:
The solution to the inequality 3x2−5x−2>0is −1
3<x<2, which
can be written in interval notation as (−1
3,2).
Question 16
Step-by-step solution: 1. Start by finding the roots of the corre-
sponding quadratic equation 2x2−7x+ 3 = 0. 2. Use the quadratic
formula x=−b±√b2−4ac
2awhere a= 2,b=−7, and c= 3 to find the roots.
3. Calculate the discriminant ∆ = b2−4ac to determine the nature
of the roots. 4. If the discriminant is positive, there are two distinct
real roots. If it is zero, there is one real root. If it is negative, there
are no real roots. 5. Once you have found the roots, plot them on a
number line to divide the number line into intervals. 6. Test a value
from each interval in the original inequality to determine the solution
set. 7. The solution to the quadratic inequality 2x2−7x+ 3 <0will
be the values that make the inequality true.Question 16: Solve the
quadratic inequality 2x2−7x+ 3 <0.
Step-by-step solution: 1. Start by finding the roots of the corre-
sponding quadratic equation 2x2−7x+ 3 = 0. 2. Use the quadratic
15
formula x=−b±√b2−4ac
2awhere a= 2,b=−7, and c= 3 to find the roots.
3. Calculate the discriminant ∆ = b2−4ac to determine the nature
of the roots. 4. If the discriminant is positive, there are two distinct
real roots. If it is zero, there is one real root. If it is negative, there
are no real roots. 5. Once you have found the roots, plot them on a
number line to divide the number line into intervals. 6. Test a value
from each interval in the original inequality to determine the solution
set. 7. The solution to the quadratic inequality 2x2−7x+ 3 <0will
be the values that make the inequality true.
Question 17
Step-by-step solution:
3
2x−1≤1
2(4x+ 7)
3x−2≤2x+7
2(Distribute on the right side)
3x−2x≤7
2+ 2 (Subtract 2xfrom both sides)
x≤11
2
Therefore, the solution to the inequality is x≤11
2.Question 17:
Solve the inequality 3
2x−1≤1
2(4x+ 7).
Step-by-step solution:
3
2x−1≤1
2(4x+ 7)
3x−2≤2x+7
2(Distribute on the right side)
3x−2x≤7
2+ 2 (Subtract 2xfrom both sides)
x≤11
2
Therefore, the solution to the inequality is x≤11
2.
Question 18
Solve the following quadratic inequality:
(x−3)(x+ 2) ≥0
Solution:
16
To solve the quadratic inequality (x−3)(x+ 2) ≥0, we need to find
the critical points where the expression equals zero and determine
the sign of the expression in each interval.
1. Find the critical points by setting the expression equal to zero:
(x−3)(x+ 2) = 0
This gives us x= 3 and x=−2as critical points.
2. Plot these critical points on a number line:
−2 3
3. Test the intervals created by the critical points by choosing test
points and determining the sign of the expression in each interval.
- For x < −2, choose x=−3:
(−3−3)(−3 + 2) = (−6)(−1) = 6 >0
Since the expression is positive, solutions in this interval are included.
- For −2<x<3, choose x= 0:
(0 −3)(0 + 2) = (−3)(2) = −6<0
Since the expression is negative, solutions in this interval are not
included.
- For x > 3, choose x= 4:
(4 −3)(4 + 2) = (1)(6) = 6 >0
Since the expression is positive, solutions in this interval are included.
4. Combine the results to write the solution set:
x∈(−∞,−2] ∪[3,∞)
Therefore, the solution to the inequality (x−3)(x+ 2) ≥0is x∈
(−∞,−2] ∪[3,∞).Question 18:
Solve the following quadratic inequality:
(x−3)(x+ 2) ≥0
Solution:
To solve the quadratic inequality (x−3)(x+ 2) ≥0, we need to find
the critical points where the expression equals zero and determine
the sign of the expression in each interval.
1. Find the critical points by setting the expression equal to zero:
(x−3)(x+ 2) = 0
This gives us x= 3 and x=−2as critical points.
17
2. Plot these critical points on a number line:
−2 3
3. Test the intervals created by the critical points by choosing test
points and determining the sign of the expression in each interval.
- For x < −2, choose x=−3:
(−3−3)(−3 + 2) = (−6)(−1) = 6 >0
Since the expression is positive, solutions in this interval are included.
- For −2<x<3, choose x= 0:
(0 −3)(0 + 2) = (−3)(2) = −6<0
Since the expression is negative, solutions in this interval are not
included.
- For x > 3, choose x= 4:
(4 −3)(4 + 2) = (1)(6) = 6 >0
Since the expression is positive, solutions in this interval are included.
4. Combine the results to write the solution set:
x∈(−∞,−2] ∪[3,∞)
Therefore, the solution to the inequality (x−3)(x+ 2) ≥0is x∈
(−∞,−2] ∪[3,∞).
Question 19
Solve the following quadratic inequality:
2x2−5x+ 2 <0
Step-by-Step Solution:
To solve the given quadratic inequality, we first rewrite it in the
standard form:
2x2−5x+ 2 <0
Next, we factor the quadratic expression:
2x2−4x−x+ 2 <0
2x(x−2) −1(x−2) <0
(2x−1)(x−2) <0
18
Now, we find the critical points by setting each factor equal to
zero:
2x−1=0⇒x=1
2
x−2 = 0 ⇒x= 2
These critical points divide the real number line into three inter-
vals: (−∞,1
2),(1
2,2),(2,+∞).
We test each interval by choosing a test point and plugging it into
the inequality to determine the sign of the expression:
For x= 0:(2(0) −1)(0 −2) = (−1)(−2) = 2 >0, which is false.
For x= 1:(2(1) −1)(1 −2) = (1)(−1) = −1<0, which is true.
For x= 3:(2(3) −1)(3 −2) = (5)(1) = 5 >0, which is false.
Therefore, the solution to the inequality 2x2−5x+ 2 <0is:
x∈1
2,2
Question 19:
Solve the following quadratic inequality:
2x2−5x+ 2 <0
Step-by-Step Solution:
To solve the given quadratic inequality, we first rewrite it in the
standard form:
2x2−5x+ 2 <0
Next, we factor the quadratic expression:
2x2−4x−x+ 2 <0
2x(x−2) −1(x−2) <0
(2x−1)(x−2) <0
Now, we find the critical points by setting each factor equal to
zero:
2x−1=0⇒x=1
2
x−2 = 0 ⇒x= 2
These critical points divide the real number line into three inter-
vals: (−∞,1
2),(1
2,2),(2,+∞).
We test each interval by choosing a test point and plugging it into
the inequality to determine the sign of the expression:
For x= 0:(2(0) −1)(0 −2) = (−1)(−2) = 2 >0, which is false.
For x= 1:(2(1) −1)(1 −2) = (1)(−1) = −1<0, which is true.
For x= 3:(2(3) −1)(3 −2) = (5)(1) = 5 >0, which is false.
19
Therefore, the solution to the inequality 2x2−5x+ 2 <0is:
x∈1
2,2
Question 20
3x2+ 4x−4>0
Step-by-step solution: 1. Begin by rewriting the inequality in
standard form:
3x2+ 4x−4>0
2. Identify the coefficients of the quadratic equation:
a= 3, b = 4, c =−4
3. Calculate the discriminant:
∆ = b2−4ac
∆=42−4(3)(−4)
∆ = 16 + 48
∆ = 64
4. Determine the nature of the roots using the discriminant: - If
∆>0, the inequality will have two distinct real roots. - If ∆=0, the
inequality will have a repeated real root. - If ∆<0, the inequality
will have no real roots.
5. Since ∆>0, the inequality will have two distinct real roots. To
find the roots, use the quadratic formula:
x=−b±√∆
2a
x=−4±√64
2(3)
x=−4±8
6
6. We get two roots:
x1=−4+8
6=4
6=2
3
x2=−4−8
6=−12
6=−2
20
7. Now, we need to determine the intervals where the inequality is
satisfied. To do this, we create a sign chart using the roots we found.
x < −2−2<x< 2
3x > 2
3
3x2+ 4x−4−+ +
8. Since the inequality is greater than zero (>0), the solution is
where the quadratic expression is positive:
x∈(−∞,−2) ∪2
3,∞
Therefore, the solution set for the quadratic inequality 3x2+ 4x−
4>0is (−∞,−2) ∪2
3,∞.Question 20: Solve the following quadratic
inequality:
3x2+ 4x−4>0
Step-by-step solution: 1. Begin by rewriting the inequality in
standard form:
3x2+ 4x−4>0
2. Identify the coefficients of the quadratic equation:
a= 3, b = 4, c =−4
3. Calculate the discriminant:
∆ = b2−4ac
∆=42−4(3)(−4)
∆ = 16 + 48
∆ = 64
4. Determine the nature of the roots using the discriminant: - If
∆>0, the inequality will have two distinct real roots. - If ∆=0, the
inequality will have a repeated real root. - If ∆<0, the inequality
will have no real roots.
5. Since ∆>0, the inequality will have two distinct real roots. To
find the roots, use the quadratic formula:
x=−b±√∆
2a
x=−4±√64
2(3)
x=−4±8
6
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6. We get two roots:
x1=−4+8
6=4
6=2
3
x2=−4−8
6=−12
6=−2
7. Now, we need to determine the intervals where the inequality is
satisfied. To do this, we create a sign chart using the roots we found.
x < −2−2<x< 2
3x > 2
3
3x2+ 4x−4−+ +
8. Since the inequality is greater than zero (>0), the solution is
where the quadratic expression is positive:
x∈(−∞,−2) ∪2
3,∞
Therefore, the solution set for the quadratic inequality 3x2+4x−4>
0is (−∞,−2) ∪2
3,∞.
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