Solve 4/x^2+36/(2-x)^2=0

$$\frac { 4 } { x ^ { 2 } } + \frac { 36 } { ( 2 - x ) ^ { 2 } } = 0$$

Solve for x (complex solution)

$$x=\frac{1}{5}+\frac{3}{5}i=0.2+0.6i$$

$$x=\frac{1}{5}-\frac{3}{5}i=0.2-0.6i$$

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}.

$$\left(x-2\right)^{2}\times 4+x^{2}\times 36=0$$

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

$$\left(x^{2}-4x+4\right)\times 4+x^{2}\times 36=0$$

Use the distributive property to multiply x^{2}-4x+4 by 4.

$$4x^{2}-16x+16+x^{2}\times 36=0$$

Combine 4x^{2} and x^{2}\times 36 to get 40x^{2}.

$$40x^{2}-16x+16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 40 for a, -16 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 40\times 16}}{2\times 40}$$

Square -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\times 40\times 16}}{2\times 40}$$

Multiply -4 times 40.

$$x=\frac{-\left(-16\right)±\sqrt{256-160\times 16}}{2\times 40}$$

Multiply -160 times 16.

$$x=\frac{-\left(-16\right)±\sqrt{256-2560}}{2\times 40}$$

Add 256 to -2560.

$$x=\frac{-\left(-16\right)±\sqrt{-2304}}{2\times 40}$$

Take the square root of -2304.

$$x=\frac{-\left(-16\right)±48i}{2\times 40}$$

The opposite of -16 is 16.

$$x=\frac{16±48i}{2\times 40}$$

Multiply 2 times 40.

$$x=\frac{16±48i}{80}$$

Now solve the equation x=\frac{16±48i}{80} when ± is plus. Add 16 to 48i.

$$x=\frac{16+48i}{80}$$

Divide 16+48i by 80.

$$x=\frac{1}{5}+\frac{3}{5}i$$

Now solve the equation x=\frac{16±48i}{80} when ± is minus. Subtract 48i from 16.

$$x=\frac{16-48i}{80}$$

Divide 16-48i by 80.

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

The equation is now solved.

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

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}.

$$\left(x-2\right)^{2}\times 4+x^{2}\times 36=0$$

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

$$\left(x^{2}-4x+4\right)\times 4+x^{2}\times 36=0$$

Use the distributive property to multiply x^{2}-4x+4 by 4.

$$4x^{2}-16x+16+x^{2}\times 36=0$$

Combine 4x^{2} and x^{2}\times 36 to get 40x^{2}.

$$40x^{2}-16x+16=0$$

Subtract 16 from both sides. Anything subtracted from zero gives its negation.

$$40x^{2}-16x=-16$$

Divide both sides by 40.

$$\frac{40x^{2}-16x}{40}=-\frac{16}{40}$$

Dividing by 40 undoes the multiplication by 40.

$$x^{2}+\left(-\frac{16}{40}\right)x=-\frac{16}{40}$$

Reduce the fraction \frac{-16}{40} to lowest terms by extracting and canceling out 8.

$$x^{2}-\frac{2}{5}x=-\frac{16}{40}$$

Reduce the fraction \frac{-16}{40} to lowest terms by extracting and canceling out 8.

$$x^{2}-\frac{2}{5}x=-\frac{2}{5}$$

Divide -\frac{2}{5}, the coefficient of the x term, by 2 to get -\frac{1}{5}. Then add the square of -\frac{1}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{2}{5}x+\left(-\frac{1}{5}\right)^{2}=-\frac{2}{5}+\left(-\frac{1}{5}\right)^{2}$$

Square -\frac{1}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=-\frac{2}{5}+\frac{1}{25}$$

Add -\frac{2}{5} to \frac{1}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=-\frac{9}{25}$$

Factor x^{2}-\frac{2}{5}x+\frac{1}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{5}\right)^{2}=-\frac{9}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{5}\right)^{2}}=\sqrt{-\frac{9}{25}}$$

Simplify.

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

Add \frac{1}{5} to both sides of the equation.

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