Solve sqrt{32x-64}=2x?

$$\sqrt { 32 x - 64 } = 2 x ?$$

Solve for x

$$x=4$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{32x-64}\right)^{2}=\left(2x\right)^{2}$$

Calculate \sqrt{32x-64} to the power of 2 and get 32x-64.

$$32x-64=\left(2x\right)^{2}$$

Expand \left(2x\right)^{2}.

$$32x-64=2^{2}x^{2}$$

Calculate 2 to the power of 2 and get 4.

$$32x-64=4x^{2}$$

Subtract 4x^{2} from both sides.

$$32x-64-4x^{2}=0$$

Divide both sides by 4.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-16. To find a and b, set up a system to be solved.

$$a+b=8 ab=-\left(-16\right)=16$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 16.

$$1,16 2,8 4,4$$

Calculate the sum for each pair.

$$1+16=17 2+8=10 4+4=8$$

The solution is the pair that gives sum 8.

$$a=4 b=4$$

Rewrite -x^{2}+8x-16 as \left(-x^{2}+4x\right)+\left(4x-16\right).

$$\left(-x^{2}+4x\right)+\left(4x-16\right)$$

Factor out -x in the first and 4 in the second group.

$$-x\left(x-4\right)+4\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(-x+4\right)$$

To find equation solutions, solve x-4=0 and -x+4=0.

$$x=4 x=4$$

Substitute 4 for x in the equation \sqrt{32x-64}=2x.

$$\sqrt{32\times 4-64}=2\times 4$$

Simplify. The value x=4 satisfies the equation.

$$8=8$$

Substitute 4 for x in the equation \sqrt{32x-64}=2x.

$$\sqrt{32\times 4-64}=2\times 4$$

Simplify. The value x=4 satisfies the equation.

$$8=8$$

List all solutions of \sqrt{32x-64}=2x.

$$x=4 x=4$$