`x^2 - 2x - 8 = 0` Solve the equation.

Textbook Question

Chapter 5, 5.2 - Problem 11 - McDougal Littell Algebra 2 (1st Edition, Ron Larson).
See all solutions for this textbook.

2 Answers | Add Yours

Educator Approved

Educator Approved
atyourservice's profile pic

atyourservice | Student, Grade 11 | (Level 3) Valedictorian

Posted on

We can solve this using the AC factoring method:

`x^2 - 2x - 8`

Multiply a by c

`1 xx -8  `

Find factors of -8 that add up to b (-2)

these numbers will be -4 and 2, plug these in as b

`x^2 - 4x + 2x - 8`

now group:

`(x^2 - 4x) + (2x - 8)`

Factor out the greatest common factors:

x(x - 4 ) + 2 (x - 4 )

Put the numbers outside in a parentheses together :

(x + 2) (x - 4)

set the numbers outside of the parentheses equal to 0

x + 2 = 0

x = -2

x - 4 = 0

x = 4

To check:

`-2^2 - 2(-2) - 8`

4 + 4 - 8

8 - 8 = 0

`4^2 - 2(4) - 8`

16 - 8 - 8

16 - 16 =0

Educator Approved

Educator Approved
iamkaori's profile picture

iamkaori | Student, Grade 9 | (Level 2) Salutatorian

Posted on

x^2 - 2x - 8 = 0

Factor them using the magic x to get:

(x - 4) (x + 2) = 0

since -4 * 2 = -8

and -4 + 2 = -2 (the coefficient of x).

So, when solved you get x = 4, -2.

Check for extraneous solutions by plugging them into the equation.

4^2 - 2 (4) - 8 = 0

= 16 - 8 - 8 =0 (correct)

(-2)^2 - 2 (-2) - 8 = 0

= 4 + 4 - 8 = 0 (correct)

So, the answer is x = 4, -2.

We’ve answered 318,957 questions. We can answer yours, too.

Ask a question