Section 1.2 Quadratic Equations

Section 1.2
Quadratic Equations

Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Quadratic Equations


Solve the equation: x − 4x = 0
2

x( x − 4) = 0

x = 0 or x + 4 = 0

x = 0 or x − 4

The solution set is { 0, 4}


Solve the equation: x2 = 6 − x

x + x−6 = 0
2

( x + 3)( x − 2) = 0

x = −3 or x = 2

The solution set is { −3, 2}


Solve each equation.
( x + 3)
2
(a) x = 7
2
(b) =9
x+3= ± 9
x=± 7
x + 3 = ±3

x = 7 or x = − 7 x + 3 = 3 or x + 3 = −3
x = 0 or x = −6
{
The solution set is − 7, 7 }
The solution set is { 0, −6}


Solve by completing the square: 2x2 + 6x − 5 = 0
2x + 6x = 5
2
3 19
x+ =±
2 4
2x2 6x 5
+ =
2 2 2 19 3
x=± −
9 5 9 2 2
x + 3x + = +
2

4 2 4 19 − 3 19 − 3
x= or −
2 2 2
 3  19
x+  =
 2 4

Use the quadratic formula to find the real solutions if any, of the equation

2x − 4x +1 = 0
2

ax + bx + c = 0
2

b 2 − 4ac = (−4) 2 − 4(2)(1) = 16 − 8 = 8

b − 4ac > 0 so there are two real solutions
2

which can be found using the quadratic formula.



2x − 4x +1 = 0
2 −b ± b − 4ac 2
x=
2a

−(−4) ± 8 4± 8 2± 2
x= = =
2(2) 4 2

2 + 2 2 − 2 
 
The solution set is  , 
 2
 2 



1 2
x − 6 x + 18 = 0 x 2 − 12 x + 36 = 0
2
ax 2 + bx + c = 0

b 2 − 4ac = (−12) 2 − 4(1)(36) = 144 − 144 = 0

b − 4ac = 0 so there is a repeated solution
2

which can be found using the quadratic formula.


1 2
x − 6 x + 18 = 0 x − 12 x + 36 = 0
2

2
ax 2 + bx + c = 0
−b ± b 2 − 4ac
x=
2a

−(−12) ± 0 12
x= = =6 The solution set is { 6}
2(1) 2



2x + 3 = 2x
2

Since b − 4ac < 0, 2

2x − 2x + 3 = 0
2
there is no real solution.
ax 2 + bx + c = 0

b − 4ac = ( −2 ) − 4(2)(3)
2 2

= 4 − 24 = −20



1 2 −1 ± 49 −1 ± 7
6+ − 2 = 0 x= =
x x 2(6) 12
6 x2 + x − 2 = 0 −1 + 7 1
= =
12 2
b − 4ac = ( 1) − 4(6)(−2)
2
2
−1 − 7 2
= =−
= 1 + 48 = 49 12 3
Since b 2 − 4ac > 0, 1 2
The solution set is  , − 
there are two real solutions. 2 3

Section 1.2 Quadratic Equations

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Section 1.2 Quadratic Equations