Intermediate Algebra (11th edition)

(Marvins-Underground-K-12) #1

Completing the Square


To solve


Step 1 If divide each side by a.


Step 2 Write the equation with the variable terms on
one side and the constant on the other.


Step 3 Take half the coefficient of xand square it.


Step 4 Add the square to each side.


Step 5 Factor the perfect square trinomial, and write it
as the square of a binomial. Simplify the other
side.


Step 6 Use the square root property to complete the
solution.


aZ1,

ax^2 +bx+c= 0 1 with aZ 02 :

CONCEPTS EXAMPLES


Solve
Divide by 2.
Add 9.

Add 1.
Factor. Add.
or Square root
or

property

The solution set is E 1 + 210 , 1- (^210) F,or E 1  (^210) F
x= 1 + 210 x = 1 - 210
x- 1 = 210 x - 1 =- 210
1 x- 122 = 10
x^2 - 2 x+ 1 = 9 + 1


C


1
2 1 -^22 D

(^2) = 1 - 122 = 1
x^2 - 2 x= 9
x^2 - 2 x- 9 = 0
2 x^2 - 4 x- 18 =0.


9.2 The Quadratic Formula


Quadratic Formula


The solutions of are
given by


The Discriminant


If a, b, and c are integers, then the discriminant,
of determines the number
and type of solutions as follows.


b^2  4 ac, ax^2 +bx+c= 0


x

b 2 b^2  4 ac
2 a

.


ax^2 +bx+c= 0 1 with aZ 02

Solve

or

The solution set is

For the discriminant is
Two rational solutions
For the discriminant is
12 - 4142112 =- 15. Two nonreal complex solutions

4 x^2 +x+ 1 =0,

32 - 41121 - 102 = 49.


x^2 + 3 x- 10 =0,

E-1, -


2
3 F.

x=

- 5 - 1


6


x= =- 1

- 5 + 1


6


=-


2


3


x=

- 5  252 - 4132122


2132


=


- 5  1


6


3 x^2 + 5 x+ 2 =0.

Number and
Discriminant Type of Solutions
Positive, the square Two rational solutions
of an integer
Positive, not the Two irrational solutions
square of an integer
Zero One rational solution
Negative Two nonreal complex solutions

9.3 Equations Quadratic in Form


A nonquadratic equation that can be written in the form


for and an algebraic expression u, is called
quadratic in form. Substitute ufor the expression, solve
for u, and then solve for the variable in the expression.


aZ 0

au^2 buc0,

Solve
Let
Factor.

or

or

or Subtract 5.

The solution set is E-7, -^163 F.

x=- x=- 7

16


3


x + 5 =- x+ 5 =- 2 x+ 5 =u

1


3


u=- u=- 2

1


3


13 u+ 121 u+ 22 = 0

3 u^2 + 7 u+ 2 = 0 u=x+5.

31 x+ 522 + 71 x+ 52 + 2 =0.

560 CHAPTER 9 Quadratic Equations, Inequalities, and Functions


(continued)
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