Solve 20/(1+x)^2=19/x^2

$$\frac { 20 } { ( 1 + x ) ^ { 2 } } = \frac { 19 } { x ^ { 2 } }$$

Solve for x

$$x=2\sqrt{95}+19\approx 38.49358869$$

$$x=19-2\sqrt{95}\approx -0.49358869$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -1,0 since division by zero is not defined. Multiply both sides of the equation by x^{2}\left(x+1\right)^{2}, the least common multiple of \left(1+x\right)^{2},x^{2}.

$$x^{2}\times 20=\left(x+1\right)^{2}\times 19$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$x^{2}\times 20=\left(x^{2}+2x+1\right)\times 19$$

Use the distributive property to multiply x^{2}+2x+1 by 19.

$$x^{2}\times 20=19x^{2}+38x+19$$

Subtract 19x^{2} from both sides.

$$x^{2}\times 20-19x^{2}=38x+19$$

Combine x^{2}\times 20 and -19x^{2} to get x^{2}.

$$x^{2}=38x+19$$

Subtract 38x from both sides.

$$x^{2}-38x=19$$

Subtract 19 from both sides.

$$x^{2}-38x-19=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -38 for b, and -19 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-38\right)±\sqrt{\left(-38\right)^{2}-4\left(-19\right)}}{2}$$

Square -38.

$$x=\frac{-\left(-38\right)±\sqrt{1444-4\left(-19\right)}}{2}$$

Multiply -4 times -19.

$$x=\frac{-\left(-38\right)±\sqrt{1444+76}}{2}$$

Add 1444 to 76.

$$x=\frac{-\left(-38\right)±\sqrt{1520}}{2}$$

Take the square root of 1520.

$$x=\frac{-\left(-38\right)±4\sqrt{95}}{2}$$

The opposite of -38 is 38.

$$x=\frac{38±4\sqrt{95}}{2}$$

Now solve the equation x=\frac{38±4\sqrt{95}}{2} when ± is plus. Add 38 to 4\sqrt{95}.

$$x=\frac{4\sqrt{95}+38}{2}$$

Divide 38+4\sqrt{95} by 2.

$$x=2\sqrt{95}+19$$

Now solve the equation x=\frac{38±4\sqrt{95}}{2} when ± is minus. Subtract 4\sqrt{95} from 38.

$$x=\frac{38-4\sqrt{95}}{2}$$

Divide 38-4\sqrt{95} by 2.

$$x=19-2\sqrt{95}$$

The equation is now solved.

$$x=2\sqrt{95}+19 x=19-2\sqrt{95}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -1,0 since division by zero is not defined. Multiply both sides of the equation by x^{2}\left(x+1\right)^{2}, the least common multiple of \left(1+x\right)^{2},x^{2}.

$$x^{2}\times 20=\left(x+1\right)^{2}\times 19$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$x^{2}\times 20=\left(x^{2}+2x+1\right)\times 19$$

Use the distributive property to multiply x^{2}+2x+1 by 19.

$$x^{2}\times 20=19x^{2}+38x+19$$

Subtract 19x^{2} from both sides.

$$x^{2}\times 20-19x^{2}=38x+19$$

Combine x^{2}\times 20 and -19x^{2} to get x^{2}.

$$x^{2}=38x+19$$

Subtract 38x from both sides.

$$x^{2}-38x=19$$

Divide -38, the coefficient of the x term, by 2 to get -19. Then add the square of -19 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-38x+\left(-19\right)^{2}=19+\left(-19\right)^{2}$$

Square -19.

$$x^{2}-38x+361=19+361$$

Add 19 to 361.

$$x^{2}-38x+361=380$$

Factor x^{2}-38x+361. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-19\right)^{2}=380$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-19\right)^{2}}=\sqrt{380}$$

Simplify.

$$x-19=2\sqrt{95} x-19=-2\sqrt{95}$$

Add 19 to both sides of the equation.

$$x=2\sqrt{95}+19 x=19-2\sqrt{95}$$