Beginning Algebra, 11th Edition

Solving Quadratic Equations of the Form

Solve each equation. Write radicals in simplified form.

(a)

By the square root property, if then

or

Checkeach solution by substituting it for xin the original equation. The solution set is

5 - 4 , 46 , or 5  46.

x= 216 = 4 x=- 216 =- 4.

x^2 =16,

x^2 = 16

EXAMPLE 2 x (^2) =k
SECTION 9.1 Solving Quadratic Equations by the Square Root Property 555
NOTE When we solve an equation, we must find allvalues of the variable that sat-
isfy the equation. Therefore, we want both the positive and negative square roots of k.
This notation indicates
twosolutions, one positive
and one negative.
(b)
The solutions are or so the solution set is
(c)
Add 32.
Divide by 5.
or Square root property
m = 222 or m=- 222 28 = 24 # 22 = 222
m= 28 m=- 28
m^2 = 8
5 m^2 = 40
5 m^2 - 32 = 8
z= 25 z= - 25 , E (^25) F.
z^2 = 5
The solution set is
(d)
Since is a negative number and since the square of a real number cannot be
negative, there is no real number solutionof this equation. (At this point, we are only
concerned with real numbersolutions.) The solution set is
(e)
Subtract 5.
Divide by 3.
or Square root property
The solution set is NOW TRY
OBJECTIVE 3 Solve equations of the form where
In each equation above, the exponent 2 appeared with a single variable as its base. We
can extend the square root property to solve equations in which the base is a binomial.
Solving Quadratic Equations of the Form
Solve each equation.
(a)
or Square root property
or
x= 7 or x=- 1 Add 3.
x - 3 = 4 x- 3 =- 4 216 = 4
x - 3 = 216 x - 3 = - 216
1 x- 322 = 16
EXAMPLE 3 1 x+b 22 =k
1 axb 22 k, k> 0.

E 22 F.

x= 22 x=- 22

x^2 = 2

3 x^2 = 6

3 x^2 + 5 = 11

0.

- 4

p^2 =- 4

E^222 F.

Don’t stop here.

Simplify the

radicals.

Use

as the base.

1 x- 32

NOW TRY
EXERCISE 2
Solve each equation. Write
radicals in simplified form.

(a)

(b)

(c)

(d) 2 x^2 - 5 = 35

x^2 =- 144

x^2 = 13

t^2 = 25

NOW TRY ANSWERS

2. (a) (b)

(c) 0 (d)E (^225) F
5  56 E 213 F