Chapter Fourteen — Quadratic Equations and Inequalities

The Humongous Book of Algebra Problems

300

Rationalize by multiplying the numerator and denominator by .

The solution to the equation is , , or .

Completing the Square

Make a trinomial into a perfect square

14.12 Given the quadratic expression x

2

– 8x + n, identify the value of n that makes

the trinomial a perfect square and verify your answer.

Divide the coefﬁcient of x by 2.

–8 ÷ 2 = –4

Square the quotient.

(–4)

2

= 16

Substituting n = 16 into the expression creates the perfect square x

2

– 8x + 16. To

verify that the quadratic is a perfect square, factor it.

x

2

– 8x + 16 = (x – 4)(x – 4) = (x – 4)

2

14.13 Given the expression x

2

– 3x + b, identify the value of b that makes the

trinomial a perfect square.

Apply the technique outlined in Problem 14.12: divide the coefﬁcient of x by 2

(that is, multiply it by ) and square the result.

14.14 Solve the equation by completing the square: x

2

– 6x = 0.

Convert the quadratic binomial x

2

– 6x into a perfect square quadratic trinomial

by dividing the coefﬁcient of x by 2 (–6 ÷ 2 = –3) and squaring the result

([–3]

2

= 9).

x

2

– 6x + 9 = 0 + 9

Notice that you cannot add 9 only to the left side of the equation. Any real

number added to one side of the equation must be added to the other side

as well.

x

2

– 6x + 9 = 9

The equation

is allowed to

have up to three

real number solutions

because the highest

power of x in the

original equation

was 3.

x

2

– 8x + 16

is a perfect

square because it’s

equal to some quantity

multiplied times itself:

(x – 4)(x – 4). It’s the

same logic that makes

25 a perfect square:

5

˙

5 = 25.

This trick only

works if you have

an x

2

-term with a

coefcient of 1

and an x-term.

Chapter Fourteen — Quadratic Equations and Inequalities

The Humongous Book of Algebra Problems

301

Factor the left side of the equation.

(x – 3)

2

= 9

Solve for x by taking the square root of both sides of the equation. Remember

that this operation requires you to include “±” immediately right of the equal

sign.

Solve both equations that result.

The solution to the equation is x = 0 or x = 6.

14.15 Solve by completing the square: x

2

+ 2x – 7 = 0.

Move the constant to the right side of the equation by adding 7 to both sides.

x

2

+ 2x = 7

Divide the coefﬁcient of x by 2 and square the result: (2 ÷ 2)

2

= 1

2

= 1. Add 1 to

both sides of the equation.

Factor the perfect square.

(x + 1)

2

= 8

Solve for x by taking the square root of both sides of the equation.

Solve for x by subtracting 1 from both sides of the equation.

The solution to the equation is or .

Here’s

a tip: When

you factor

the trinomial,

you’ll always get

(x + something)

2

.

That “something” is

half of the original

x-coefcient, the

number you

squared to get 9

a few steps

ago.

The equation

x – 3 = ±3

represents two

different equations:

x – 3 = +3 or

x – 3 = –3.

This leaves

an x

2

-term

and an x-term

on the left side of

the equation, so you

can come up with a

constant to make

a perfect square

like in Problems

14.12–14.14.

Chapter Fourteen — Quadratic Equations and Inequalities

The Humongous Book of Algebra Problems

302

14.16 Solve by completing the square: x

2

– 5x + 1 = 0.

Move the constant to the right side of the equation by subtracting 1 from both

sides.

x

2

– 5x = –1

Square half of the x-coefﬁcient: ; add the result to both

sides of the equation.

Factor the left side of the equation.

Solve the equation for x.

The solution to the equation is or .

Don’t forget

the factoring

trick: this number

is always half of

the original x-

coefcient.

You can

add these

fractions because

they have the same

denominator.

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