Solve for x x^2-14x+49=0

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

Step 1

Factor using the perfect square rule.

Rewrite $49$ as $7^{2}$ .

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

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

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

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 7$ .

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

Step 2

Set the $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 3

Add $7$ to both sides of the equation.

$x = 7$