Solve (100+4x^2)divx=8x

$$(100+4 { x }^{ 2 } ) \div x=8x$$

Solve for x

$$x=5$$

$$x=-5$$

Steps by Finding Square Root

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$100+4x^{2}=8xx$$

Multiply x and x to get x^{2}.

$$100+4x^{2}=8x^{2}$$

Subtract 8x^{2} from both sides.

$$100+4x^{2}-8x^{2}=0$$

Combine 4x^{2} and -8x^{2} to get -4x^{2}.

$$100-4x^{2}=0$$

Subtract 100 from both sides. Anything subtracted from zero gives its negation.

$$-4x^{2}=-100$$

Divide both sides by -4.

$$x^{2}=\frac{-100}{-4}$$

Divide -100 by -4 to get 25.

$$x^{2}=25$$

Take the square root of both sides of the equation.

$$x=5 x=-5$$

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 x.

$$100+4x^{2}=8xx$$

Multiply x and x to get x^{2}.

$$100+4x^{2}=8x^{2}$$

Subtract 8x^{2} from both sides.

$$100+4x^{2}-8x^{2}=0$$

Combine 4x^{2} and -8x^{2} to get -4x^{2}.

$$100-4x^{2}=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$-4x^{2}+100=0$$

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

$$x=\frac{0±\sqrt{0^{2}-4\left(-4\right)\times 100}}{2\left(-4\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-4\right)\times 100}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{0±\sqrt{16\times 100}}{2\left(-4\right)}$$

Multiply 16 times 100.

$$x=\frac{0±\sqrt{1600}}{2\left(-4\right)}$$

Take the square root of 1600.

$$x=\frac{0±40}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{0±40}{-8}$$

Now solve the equation x=\frac{0±40}{-8} when ± is plus. Divide 40 by -8.

$$x=-5$$

Now solve the equation x=\frac{0±40}{-8} when ± is minus. Divide -40 by -8.

$$x=5$$

The equation is now solved.

$$x=-5 x=5$$