Solve 6/x^2=32/(2-x)^2

$$\frac { 6 } { x ^ { 2 } } = \frac { 32 } { ( 2 - x ) ^ { 2 } }$$

Solve for x

$$x=\frac{8\sqrt{3}-6}{13}\approx 0.604338959$$

$$x=\frac{-8\sqrt{3}-6}{13}\approx -1.527415882$$

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}.

$$\left(x-2\right)^{2}\times 6=x^{2}\times 32$$

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

$$\left(x^{2}-4x+4\right)\times 6=x^{2}\times 32$$

Use the distributive property to multiply x^{2}-4x+4 by 6.

$$6x^{2}-24x+24=x^{2}\times 32$$

Subtract x^{2}\times 32 from both sides.

$$6x^{2}-24x+24-x^{2}\times 32=0$$

Combine 6x^{2} and -x^{2}\times 32 to get -26x^{2}.

$$-26x^{2}-24x+24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -26 for a, -24 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\left(-26\right)\times 24}}{2\left(-26\right)}$$

Square -24.

$$x=\frac{-\left(-24\right)±\sqrt{576-4\left(-26\right)\times 24}}{2\left(-26\right)}$$

Multiply -4 times -26.

$$x=\frac{-\left(-24\right)±\sqrt{576+104\times 24}}{2\left(-26\right)}$$

Multiply 104 times 24.

$$x=\frac{-\left(-24\right)±\sqrt{576+2496}}{2\left(-26\right)}$$

Add 576 to 2496.

$$x=\frac{-\left(-24\right)±\sqrt{3072}}{2\left(-26\right)}$$

Take the square root of 3072.

$$x=\frac{-\left(-24\right)±32\sqrt{3}}{2\left(-26\right)}$$

The opposite of -24 is 24.

$$x=\frac{24±32\sqrt{3}}{2\left(-26\right)}$$

Multiply 2 times -26.

$$x=\frac{24±32\sqrt{3}}{-52}$$

Now solve the equation x=\frac{24±32\sqrt{3}}{-52} when ± is plus. Add 24 to 32\sqrt{3}.

$$x=\frac{32\sqrt{3}+24}{-52}$$

Divide 24+32\sqrt{3} by -52.

$$x=\frac{-8\sqrt{3}-6}{13}$$

Now solve the equation x=\frac{24±32\sqrt{3}}{-52} when ± is minus. Subtract 32\sqrt{3} from 24.

$$x=\frac{24-32\sqrt{3}}{-52}$$

Divide 24-32\sqrt{3} by -52.

$$x=\frac{8\sqrt{3}-6}{13}$$

The equation is now solved.

$$x=\frac{-8\sqrt{3}-6}{13} x=\frac{8\sqrt{3}-6}{13}$$

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}.

$$\left(x-2\right)^{2}\times 6=x^{2}\times 32$$

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

$$\left(x^{2}-4x+4\right)\times 6=x^{2}\times 32$$

Use the distributive property to multiply x^{2}-4x+4 by 6.

$$6x^{2}-24x+24=x^{2}\times 32$$

Subtract x^{2}\times 32 from both sides.

$$6x^{2}-24x+24-x^{2}\times 32=0$$

Combine 6x^{2} and -x^{2}\times 32 to get -26x^{2}.

$$-26x^{2}-24x+24=0$$

Subtract 24 from both sides. Anything subtracted from zero gives its negation.

$$-26x^{2}-24x=-24$$

Divide both sides by -26.

$$\frac{-26x^{2}-24x}{-26}=-\frac{24}{-26}$$

Dividing by -26 undoes the multiplication by -26.

$$x^{2}+\left(-\frac{24}{-26}\right)x=-\frac{24}{-26}$$

Reduce the fraction \frac{-24}{-26} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{12}{13}x=-\frac{24}{-26}$$

Reduce the fraction \frac{-24}{-26} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{12}{13}x=\frac{12}{13}$$

Divide \frac{12}{13}, the coefficient of the x term, by 2 to get \frac{6}{13}. Then add the square of \frac{6}{13} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{12}{13}x+\left(\frac{6}{13}\right)^{2}=\frac{12}{13}+\left(\frac{6}{13}\right)^{2}$$

Square \frac{6}{13} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{12}{13}x+\frac{36}{169}=\frac{12}{13}+\frac{36}{169}$$

Add \frac{12}{13} to \frac{36}{169} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{12}{13}x+\frac{36}{169}=\frac{192}{169}$$

Factor x^{2}+\frac{12}{13}x+\frac{36}{169}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{6}{13}\right)^{2}=\frac{192}{169}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{6}{13}\right)^{2}}=\sqrt{\frac{192}{169}}$$

Simplify.

$$x+\frac{6}{13}=\frac{8\sqrt{3}}{13} x+\frac{6}{13}=-\frac{8\sqrt{3}}{13}$$

Subtract \frac{6}{13} from both sides of the equation.

$$x=\frac{8\sqrt{3}-6}{13} x=\frac{-8\sqrt{3}-6}{13}$$