# How do you solve x^2 = 14x – 49?

Jul 8, 2015

The answer is $x = 7$.

#### Explanation:

${x}^{2} = 14 x - 49$

Subtract $14 x$ from both sides of the equation.

${x}^{2} - 14 x = - 49$

Add $49$ to both sides of the equation.

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

${x}^{2} - 14 x + 49$ is a perfect square trinomial in the form ${a}^{2} + 2 a b + {b}^{2} = {\left(a + b\right)}^{2}$, where $a = x \mathmr{and} b = - 7$.

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

${\left(x - 7\right)}^{2} = 0$

Take the square root of both sides.

$\left(x - 7\right) = \sqrt{0}$ =

$x - 7 = 0$ =

$x = 7$