Find All Complex Solutions x^2-8x+16=0

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

Step 1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2

Substitute the values $a = 1$ , $b = - 8$ , and $c = 16$ into the quadratic formula and solve for $x$ .

$\frac{8 \pm \sqrt{{(- 8)}^{2} - 4 \cdot (1 \cdot 16)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 8$ to the power of $2$ .

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 1 \cdot 16}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 16$ .

Multiply $- 4$ by $1$ .

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 16}}{2 \cdot 1}$

Multiply $- 4$ by $16$ .

$x = \frac{8 \pm \sqrt{64 - 64}}{2 \cdot 1}$

Subtract $64$ from $64$ .

$x = \frac{8 \pm \sqrt{0}}{2 \cdot 1}$

Rewrite $0$ as $0^{2}$ .

$x = \frac{8 \pm \sqrt{0^{2}}}{2 \cdot 1}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{8 \pm 0}{2 \cdot 1}$

$8$ plus or minus $0$ is $8$ .

$x = \frac{8}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{8}{2}$

Divide $8$ by $2$ .

$x = 4$

Step 4

The final answer is the combination of both solutions.

$x = 4$ Double roots