Solve 64=left(2-2xright)^2/x^2

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

Solve for x

$$x=-\frac{1}{3}\approx -0.333333333$$

$$x=\frac{1}{5}=0.2$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-2x\right)^{2}.

$$64x^{2}=4-8x+4x^{2}$$

Subtract 4 from both sides.

$$64x^{2}-4=-8x+4x^{2}$$

Add 8x to both sides.

$$64x^{2}-4+8x=4x^{2}$$

Subtract 4x^{2} from both sides.

$$64x^{2}-4+8x-4x^{2}=0$$

Combine 64x^{2} and -4x^{2} to get 60x^{2}.

$$60x^{2}-4+8x=0$$

Divide both sides by 4.

$$15x^{2}-1+2x=0$$

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

$$15x^{2}+2x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 15x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=2 ab=15\left(-1\right)=-15$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -15.

$$-1,15 -3,5$$

Calculate the sum for each pair.

$$-1+15=14 -3+5=2$$

The solution is the pair that gives sum 2.

$$a=-3 b=5$$

Rewrite 15x^{2}+2x-1 as \left(15x^{2}-3x\right)+\left(5x-1\right).

$$\left(15x^{2}-3x\right)+\left(5x-1\right)$$

Factor out 3x in 15x^{2}-3x.

$$3x\left(5x-1\right)+5x-1$$

Factor out common term 5x-1 by using distributive property.

$$\left(5x-1\right)\left(3x+1\right)$$

To find equation solutions, solve 5x-1=0 and 3x+1=0.

$$x=\frac{1}{5} x=-\frac{1}{3}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-2x\right)^{2}.

$$64x^{2}=4-8x+4x^{2}$$

Subtract 4 from both sides.

$$64x^{2}-4=-8x+4x^{2}$$

Add 8x to both sides.

$$64x^{2}-4+8x=4x^{2}$$

Subtract 4x^{2} from both sides.

$$64x^{2}-4+8x-4x^{2}=0$$

Combine 64x^{2} and -4x^{2} to get 60x^{2}.

$$60x^{2}-4+8x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$60x^{2}+8x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 60 for a, 8 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 60\left(-4\right)}}{2\times 60}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 60\left(-4\right)}}{2\times 60}$$

Multiply -4 times 60.

$$x=\frac{-8±\sqrt{64-240\left(-4\right)}}{2\times 60}$$

Multiply -240 times -4.

$$x=\frac{-8±\sqrt{64+960}}{2\times 60}$$

Add 64 to 960.

$$x=\frac{-8±\sqrt{1024}}{2\times 60}$$

Take the square root of 1024.

$$x=\frac{-8±32}{2\times 60}$$

Multiply 2 times 60.

$$x=\frac{-8±32}{120}$$

Now solve the equation x=\frac{-8±32}{120} when ± is plus. Add -8 to 32.

$$x=\frac{24}{120}$$

Reduce the fraction \frac{24}{120} to lowest terms by extracting and canceling out 24.

$$x=\frac{1}{5}$$

Now solve the equation x=\frac{-8±32}{120} when ± is minus. Subtract 32 from -8.

$$x=-\frac{40}{120}$$

Reduce the fraction \frac{-40}{120} to lowest terms by extracting and canceling out 40.

$$x=-\frac{1}{3}$$

The equation is now solved.

$$x=\frac{1}{5} x=-\frac{1}{3}$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-2x\right)^{2}.

$$64x^{2}=4-8x+4x^{2}$$

Add 8x to both sides.

$$64x^{2}+8x=4+4x^{2}$$

Subtract 4x^{2} from both sides.

$$64x^{2}+8x-4x^{2}=4$$

Combine 64x^{2} and -4x^{2} to get 60x^{2}.

$$60x^{2}+8x=4$$

Divide both sides by 60.

$$\frac{60x^{2}+8x}{60}=\frac{4}{60}$$

Dividing by 60 undoes the multiplication by 60.

$$x^{2}+\frac{8}{60}x=\frac{4}{60}$$

Reduce the fraction \frac{8}{60} to lowest terms by extracting and canceling out 4.

$$x^{2}+\frac{2}{15}x=\frac{4}{60}$$

Reduce the fraction \frac{4}{60} to lowest terms by extracting and canceling out 4.

$$x^{2}+\frac{2}{15}x=\frac{1}{15}$$

Divide \frac{2}{15}, the coefficient of the x term, by 2 to get \frac{1}{15}. Then add the square of \frac{1}{15} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{15}x+\left(\frac{1}{15}\right)^{2}=\frac{1}{15}+\left(\frac{1}{15}\right)^{2}$$

Square \frac{1}{15} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{15}x+\frac{1}{225}=\frac{1}{15}+\frac{1}{225}$$

Add \frac{1}{15} to \frac{1}{225} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{15}x+\frac{1}{225}=\frac{16}{225}$$

Factor x^{2}+\frac{2}{15}x+\frac{1}{225}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{15}\right)^{2}=\frac{16}{225}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{15}\right)^{2}}=\sqrt{\frac{16}{225}}$$

Simplify.

$$x+\frac{1}{15}=\frac{4}{15} x+\frac{1}{15}=-\frac{4}{15}$$

Subtract \frac{1}{15} from both sides of the equation.

$$x=\frac{1}{5} x=-\frac{1}{3}$$