Solve 8=x+12/x

$$8=x+ \frac{ 12 }{ x }$$

Solve for x

$$x=2$$

$$x=6$$

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.

$$8x=xx+12$$

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

$$8x=x^{2}+12$$

Subtract x^{2} from both sides.

$$8x-x^{2}=12$$

Subtract 12 from both sides.

$$8x-x^{2}-12=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-x^{2}+8x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 8 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-1\right)\left(-12\right)}}{2\left(-1\right)}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-1\right)\left(-12\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-8±\sqrt{64+4\left(-12\right)}}{2\left(-1\right)}$$

Multiply 4 times -12.

$$x=\frac{-8±\sqrt{64-48}}{2\left(-1\right)}$$

Add 64 to -48.

$$x=\frac{-8±\sqrt{16}}{2\left(-1\right)}$$

Take the square root of 16.

$$x=\frac{-8±4}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-8±4}{-2}$$

Now solve the equation x=\frac{-8±4}{-2} when ± is plus. Add -8 to 4.

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

Divide -4 by -2.

$$x=2$$

Now solve the equation x=\frac{-8±4}{-2} when ± is minus. Subtract 4 from -8.

$$x=-\frac{12}{-2}$$

Divide -12 by -2.

$$x=6$$

The equation is now solved.

$$x=2 x=6$$

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 x.

$$8x=xx+12$$

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

$$8x=x^{2}+12$$

Subtract x^{2} from both sides.

$$8x-x^{2}=12$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-x^{2}+8x=12$$

Divide both sides by -1.

$$\frac{-x^{2}+8x}{-1}=\frac{12}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{8}{-1}x=\frac{12}{-1}$$

Divide 8 by -1.

$$x^{2}-8x=\frac{12}{-1}$$

Divide 12 by -1.

$$x^{2}-8x=-12$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=-12+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=-12+16$$

Add -12 to 16.

$$x^{2}-8x+16=4$$

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{4}$$

Simplify.

$$x-4=2 x-4=-2$$

Add 4 to both sides of the equation.

$$x=6 x=2$$