Pre-Algebra Examples

x^{2} + 6 x = 16

$x^{2} + 6 x = 16$

Step 1

Subtract

16

$16$ from both sides of the equation.

x^{2} + 6 x - 16 = 0

$x^{2} + 6 x - 16 = 0$

Step 2

Factor

x^{2} + 6 x - 16

$x^{2} + 6 x - 16$ using the AC method.

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Step 2.1

Consider the form

x^{2} + b x + c

$x^{2} + b x + c$ . Find a pair of integers whose product is

c

$c$ and whose sum is

b

$b$ . In this case, whose product is

- 16

$- 16$ and whose sum is

6

$6$ .

- 2, 8

$- 2, 8$

Step 2.2

Write the factored form using these integers.

(x - 2) (x + 8) = 0

$(x - 2) (x + 8) = 0$

(x - 2) (x + 8) = 0

$(x - 2) (x + 8) = 0$

Step 3

If any individual factor on the left side of the equation is equal to

0

$0$ , the entire expression will be equal to

0

$0$ .

x - 2 = 0

$x - 2 = 0$

x + 8 = 0

$x + 8 = 0$

Step 4

Set

x - 2

$x - 2$ equal to

0

$0$ and solve for

x

$x$ .

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Step 4.1

Set

x - 2

$x - 2$ equal to

0

$0$ .

x - 2 = 0

$x - 2 = 0$

Step 4.2

Add

2

$2$ to both sides of the equation.

x = 2

$x = 2$

x = 2

$x = 2$

Step 5

Set

x + 8

$x + 8$ equal to

0

$0$ and solve for

x

$x$ .

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Step 5.1

Set

x + 8

$x + 8$ equal to

0

$0$ .

x + 8 = 0

$x + 8 = 0$

Step 5.2

Subtract

8

$8$ from both sides of the equation.

x = - 8

$x = - 8$

x = - 8

$x = - 8$

Step 6

The final solution is all the values that make

(x - 2) (x + 8) = 0

$(x - 2) (x + 8) = 0$ true.

x = 2, - 8

$x = 2, - 8$