Find the Roots (Zeros) y=x^2-16x+64

$y = x^{2} - 16 x + 64$

Step 1

Set $x^{2} - 16 x + 64$ equal to $0$ .

$x^{2} - 16 x + 64 = 0$

Step 2

Solve for $x$ .

Factor using the perfect square rule.

Rewrite $64$ as $8^{2}$ .

$x^{2} - 16 x + 8^{2} = 0$

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

$16 x = 2 \cdot x \cdot 8$

Rewrite the polynomial.

$x^{2} - 2 \cdot x \cdot 8 + 8^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 8$ .

${(x - 8)}^{2} = 0$

Set the $x - 8$ equal to $0$ .

$x - 8 = 0$

Add $8$ to both sides of the equation.

$x = 8$

Step 3