Pre-Algebra Examples

x^{2} + 14 x + 49 = 0

$x^{2} + 14 x + 49 = 0$

Step 1

Factor using the perfect square rule.

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Step 1.1

Rewrite

49

$49$ as

7^{2}

$7^{2}$ .

x^{2} + 14 x + 7^{2} = 0

$x^{2} + 14 x + 7^{2} = 0$

Step 1.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

14 x = 2 \cdot x \cdot 7

$14 x = 2 \cdot x \cdot 7$

Step 1.3

Rewrite the polynomial.

x^{2} + 2 \cdot x \cdot 7 + 7^{2} = 0

$x^{2} + 2 \cdot x \cdot 7 + 7^{2} = 0$

Step 1.4

Factor using the perfect square trinomial rule

a^{2} + 2 a b + b^{2} = {(a + b)}^{2}

$a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where

a = x

$a = x$ and

b = 7

$b = 7$ .

{(x + 7)}^{2} = 0

${(x + 7)}^{2} = 0$

{(x + 7)}^{2} = 0

${(x + 7)}^{2} = 0$

Step 2

Set the

x + 7

$x + 7$ equal to

0

$0$ .

x + 7 = 0

$x + 7 = 0$

Step 3

Subtract

7

$7$ from both sides of the equation.

x = - 7

$x = - 7$

x^{2} + 14 x + 49 = 0

$x^{2} + 14 x + 49 = 0$

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