Solve 4x^2+16/5x=x

$$\frac { 4 x ^ { 2 } + 16 } { 5 x } = x$$

Solve for x

$$x=4$$

$$x=-4$$

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 5x.

$$4x^{2}+16=x\times 5x$$

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

$$4x^{2}+16=x^{2}\times 5$$

Subtract x^{2}\times 5 from both sides.

$$4x^{2}+16-x^{2}\times 5=0$$

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

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

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

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

Divide both sides by -1.

$$x^{2}=\frac{-16}{-1}$$

Fraction \frac{-16}{-1} can be simplified to 16 by removing the negative sign from both the numerator and the denominator.

$$x^{2}=16$$

Take the square root of both sides of the equation.

$$x=4 x=-4$$

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 5x.

$$4x^{2}+16=x\times 5x$$

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

$$4x^{2}+16=x^{2}\times 5$$

Subtract x^{2}\times 5 from both sides.

$$4x^{2}+16-x^{2}\times 5=0$$

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

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 0 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-1\right)\times 16}}{2\left(-1\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-1\right)\times 16}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{0±\sqrt{4\times 16}}{2\left(-1\right)}$$

Multiply 4 times 16.

$$x=\frac{0±\sqrt{64}}{2\left(-1\right)}$$

Take the square root of 64.

$$x=\frac{0±8}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{0±8}{-2}$$

Now solve the equation x=\frac{0±8}{-2} when ± is plus. Divide 8 by -2.

$$x=-4$$

Now solve the equation x=\frac{0±8}{-2} when ± is minus. Divide -8 by -2.

$$x=4$$

The equation is now solved.

$$x=-4 x=4$$