Find All Complex Solutions (2x)^2-64=0

${(2 x)}^{2} - 64 = 0$

Step 1

Simplify the left side.

Simplify each term.

Apply the product rule to $2 x$ .

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

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

$4 x^{2} - 64 = 0$

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = 4$ , $b = 0$ , and $c = - 64$ into the quadratic formula and solve for $x$ .

$\frac{0 \pm \sqrt{0^{2} - 4 \cdot (4 \cdot - 64)}}{2 \cdot 4}$

Step 4

Simplify.

Simplify the numerator.

Raising $0$ to any positive power yields $0$ .

$x = \frac{0 \pm \sqrt{0 - 4 \cdot 4 \cdot - 64}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot - 64$ .

Multiply $- 4$ by $4$ .

$x = \frac{0 \pm \sqrt{0 - 16 \cdot - 64}}{2 \cdot 4}$

Multiply $- 16$ by $- 64$ .

$x = \frac{0 \pm \sqrt{0 + 1024}}{2 \cdot 4}$

Add $0$ and $1024$ .

$x = \frac{0 \pm \sqrt{1024}}{2 \cdot 4}$

Rewrite $1024$ as $32^{2}$ .

$x = \frac{0 \pm \sqrt{32^{2}}}{2 \cdot 4}$

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

$x = \frac{0 \pm 32}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{0 \pm 32}{8}$

Simplify $\frac{0 \pm 32}{8}$ .

$x = \pm 4$

Step 5

The final answer is the combination of both solutions.

$x = 4, - 4$