Solve 5/4=left(x+5right)^2/x^2

$$\frac{ 5 }{ 4 } = \frac{ { \left(x+5 \right) }^{ 2 } }{ { x }^{ 2 } }$$

Solve for x

$$x=10\sqrt{5}+20\approx 42.360679775$$

$$x=20-10\sqrt{5}\approx -2.360679775$$

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 4x^{2}, the least common multiple of 4,x^{2}.

$$5x^{2}=4\left(x+5\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$5x^{2}=4\left(x^{2}+10x+25\right)$$

Use the distributive property to multiply 4 by x^{2}+10x+25.

$$5x^{2}=4x^{2}+40x+100$$

Subtract 4x^{2} from both sides.

$$5x^{2}-4x^{2}=40x+100$$

Combine 5x^{2} and -4x^{2} to get x^{2}.

$$x^{2}=40x+100$$

Subtract 40x from both sides.

$$x^{2}-40x=100$$

Subtract 100 from both sides.

$$x^{2}-40x-100=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -40 for b, and -100 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\left(-100\right)}}{2}$$

Square -40.

$$x=\frac{-\left(-40\right)±\sqrt{1600-4\left(-100\right)}}{2}$$

Multiply -4 times -100.

$$x=\frac{-\left(-40\right)±\sqrt{1600+400}}{2}$$

Add 1600 to 400.

$$x=\frac{-\left(-40\right)±\sqrt{2000}}{2}$$

Take the square root of 2000.

$$x=\frac{-\left(-40\right)±20\sqrt{5}}{2}$$

The opposite of -40 is 40.

$$x=\frac{40±20\sqrt{5}}{2}$$

Now solve the equation x=\frac{40±20\sqrt{5}}{2} when ± is plus. Add 40 to 20\sqrt{5}.

$$x=\frac{20\sqrt{5}+40}{2}$$

Divide 40+20\sqrt{5} by 2.

$$x=10\sqrt{5}+20$$

Now solve the equation x=\frac{40±20\sqrt{5}}{2} when ± is minus. Subtract 20\sqrt{5} from 40.

$$x=\frac{40-20\sqrt{5}}{2}$$

Divide 40-20\sqrt{5} by 2.

$$x=20-10\sqrt{5}$$

The equation is now solved.

$$x=10\sqrt{5}+20 x=20-10\sqrt{5}$$

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 4x^{2}, the least common multiple of 4,x^{2}.

$$5x^{2}=4\left(x+5\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$5x^{2}=4\left(x^{2}+10x+25\right)$$

Use the distributive property to multiply 4 by x^{2}+10x+25.

$$5x^{2}=4x^{2}+40x+100$$

Subtract 4x^{2} from both sides.

$$5x^{2}-4x^{2}=40x+100$$

Combine 5x^{2} and -4x^{2} to get x^{2}.

$$x^{2}=40x+100$$

Subtract 40x from both sides.

$$x^{2}-40x=100$$

Divide -40, the coefficient of the x term, by 2 to get -20. Then add the square of -20 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-40x+\left(-20\right)^{2}=100+\left(-20\right)^{2}$$

Square -20.

$$x^{2}-40x+400=100+400$$

Add 100 to 400.

$$x^{2}-40x+400=500$$

Factor x^{2}-40x+400. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-20\right)^{2}=500$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-20\right)^{2}}=\sqrt{500}$$

Simplify.

$$x-20=10\sqrt{5} x-20=-10\sqrt{5}$$

Add 20 to both sides of the equation.

$$x=10\sqrt{5}+20 x=20-10\sqrt{5}$$