Problems 18 through 20: Consider the quadratic equation
18. For what value of c does the equation have one real root? (Hint: Using the values of a and b found in the equation, solve D = 0 for c.)
19. For what range of values does the equation have two real roots? (Hint: The answer is an inequality. Don't simply offer examples of c that satisfy the requirement.)
20. For what range of values does the equation have two complex roots?
2
4x25x0
-=
5
x
2
=±
2
x12x360
-+=
2
x14x450
++=
16
x8
x
+=
2
3x21x0
+=
22
2
2
2
22
2
2
2
General procedure:
Given axbxc, rewrite as xbxc.
bbb
Find . Square it: .
224
b
Add to both sides of the equation.
4
bb
xbxc
44
bb
xc
24
+++=-
æö
æöæö
=
ç÷
ç÷ç÷
ç÷
èøèø
èø
æö
ç÷
ç÷
èø
æöæö
++=-+
ç÷ç÷
ç÷ç÷
èøèø
æö
=+=-+
ç÷
èø
.
Take the root of both sides and solve fo
r values of x.
æö
ç÷
ç÷
èø
22
2
2
x(x+10)=11. Put into standard form.
x10x11. Half of 10 is 5; 525. Add to bo
th sides.
x10x251125
(x5)36
x5366
x65
x651
or
x6511
ANS: x=1 or -11
Checking:
x(x+10)=11
1(1+10)=1(11)
+==
++=+
+=
+==±
=±-
=-=
=--=-
=11
(11)(-11+10)=(-11)(-1)=11
-
2
x4x30
++=
x1,3
=--
2
x5x10
+-=
2
2x7x150
+-=
2
1
x7x16
2
-=
x(x5)4
-=-
2
2
2
2
General form of a quadratic equation: a
xbxc0.
Step 1: Identify a, b, c; b or c may =
0.
x3x0; a=1, b=-3, c=0
Quadratic formula:
(3)(3)4(1)(0)
bb4ac
x=
2a2(1)
3933
22
336
x3
22
or
3
x
++=
-=
--±--
-±-
=
±±
==
+
===
-
=
2
2
2
30
0.
22
ANS: x= 3 or x=0
Checking:
x3x0
33(3)990
0(0)0
==
-=
-=-=
-=
2
2
2
2
2
x10x30; a=1, b=10, c=-3
Quadratic formula:
(10)(10)4(1)(3)
bb4ac
x=
2a2(1)
104(28)
101001210112
222
10228
528
2
ANS: x= -528
Checking:
x10x30
(-5-28)10(-5-28)30
2510282850
+-=
-±--
-±-
=
-±
-±+-±
===
-±
==-±
±
+-=
+-=
++-
102830
(2528503)(10281028)0
00
Similarly for -5+28.
--=
+--+-=
=
2
x90
-=
x3
=±
2
x90
+=
2
x3x40
-+=
2
2x3x60
++=
(
)
2
5x1130
++=
b
2a
-
æö
ç÷
èø
2
2
x20 a=1, b=0, c=2.
D=b4ac04(1)(2)8.
Since D is negative, the equation has tw
o complex roots.
+=®
-=-=-
x(x3)2
+=
2
x40.
(x2)(x2)0.
The two factors are (x+2) and (x-2).
If EITHER is equal to zero, the correspo
nding value of x is a root.
Solving the two factors for x:
(x+2)=0x=-2
(x-2)=0x2
The two roots (solutio
-=
+-=
®
®=
2
2
ns) are x=2 and x=-2.
ANS: x=2
Checking:
(2)4440
(-2)4440
±
-=-=
-=-=
2x(x2)2
+=-
2
x4xc0.
++=
2
2
22
xx6.
xx60. (Must be in this form.)
(x+2)(x-3)=0
The two factors are (x+2) and (x-3). Ch
ecking:
(x+2)(x-3)=x3x2x6xx6.
If EITHER is equal to zero, the correspo
nding value of x is a root.
Solv
-=
--=
-+-=--
2
2
2
ing the two factors for x:
(x+2)=0x=-2
(x-3)=0x3
The two roots (solutions) are x=-2 and x
=3
ANS: x=-2 or +3.
Checking:
(x)x6
(-2)(2)426
(3)(3)936
®
®=
-=
--=+=
-=-=