Solve 1/x^2=2/(2-x)^2

$$\frac { 1 } { x ^ { 2 } } = \frac { 2 } { ( 2 - x ) ^ { 2 } }$$

Solve for x

$$x=2\sqrt{2}-2\approx 0.828427125$$

$$x=-2\sqrt{2}-2\approx -4.828427125$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,2 since division by zero is not defined. Multiply both sides of the equation by x^{2}\left(x-2\right)^{2}, the least common multiple of x^{2},\left(2-x\right)^{2}.

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

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

$$x^{2}-4x+4=x^{2}\times 2$$

Subtract x^{2}\times 2 from both sides.

$$x^{2}-4x+4-x^{2}\times 2=0$$

Combine x^{2} and -x^{2}\times 2 to get -x^{2}.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -4 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-4\right)±\sqrt{16+4\times 4}}{2\left(-1\right)}$$

Multiply 4 times 4.

$$x=\frac{-\left(-4\right)±\sqrt{16+16}}{2\left(-1\right)}$$

Add 16 to 16.

$$x=\frac{-\left(-4\right)±\sqrt{32}}{2\left(-1\right)}$$

Take the square root of 32.

$$x=\frac{-\left(-4\right)±4\sqrt{2}}{2\left(-1\right)}$$

The opposite of -4 is 4.

$$x=\frac{4±4\sqrt{2}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{4±4\sqrt{2}}{-2}$$

Now solve the equation x=\frac{4±4\sqrt{2}}{-2} when ± is plus. Add 4 to 4\sqrt{2}.

$$x=\frac{4\sqrt{2}+4}{-2}$$

Divide 4+4\sqrt{2} by -2.

$$x=-2\sqrt{2}-2$$

Now solve the equation x=\frac{4±4\sqrt{2}}{-2} when ± is minus. Subtract 4\sqrt{2} from 4.

$$x=\frac{4-4\sqrt{2}}{-2}$$

Divide 4-4\sqrt{2} by -2.

$$x=2\sqrt{2}-2$$

The equation is now solved.

$$x=-2\sqrt{2}-2 x=2\sqrt{2}-2$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,2 since division by zero is not defined. Multiply both sides of the equation by x^{2}\left(x-2\right)^{2}, the least common multiple of x^{2},\left(2-x\right)^{2}.

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

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

$$x^{2}-4x+4=x^{2}\times 2$$

Subtract x^{2}\times 2 from both sides.

$$x^{2}-4x+4-x^{2}\times 2=0$$

Combine x^{2} and -x^{2}\times 2 to get -x^{2}.

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

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by -1.

$$\frac{-x^{2}-4x}{-1}=-\frac{4}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{4}{-1}\right)x=-\frac{4}{-1}$$

Divide -4 by -1.

$$x^{2}+4x=-\frac{4}{-1}$$

Divide -4 by -1.

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

Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square 2.

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

Add 4 to 4.

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

Factor x^{2}+4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+2\right)^{2}=8$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{8}$$

Simplify.

$$x+2=2\sqrt{2} x+2=-2\sqrt{2}$$

Subtract 2 from both sides of the equation.

$$x=2\sqrt{2}-2 x=-2\sqrt{2}-2$$