Solve 8/x+2=x

$$\frac { 8 } { x } + 2 = x$$

Solve for x

$$x=-2$$

$$x=4$$

Steps Using Factoring By Grouping

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

$$8+x\times 2=xx$$

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

$$8+x\times 2=x^{2}$$

Subtract x^{2} from both sides.

$$8+x\times 2-x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+8. To find a and b, set up a system to be solved.

$$a+b=2 ab=-8=-8$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -8.

$$-1,8 -2,4$$

Calculate the sum for each pair.

$$-1+8=7 -2+4=2$$

The solution is the pair that gives sum 2.

$$a=4 b=-2$$

Rewrite -x^{2}+2x+8 as \left(-x^{2}+4x\right)+\left(-2x+8\right).

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

Factor out -x in the first and -2 in the second group.

$$-x\left(x-4\right)-2\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(-x-2\right)$$

To find equation solutions, solve x-4=0 and -x-2=0.

$$x=4 x=-2$$

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.

$$8+x\times 2=xx$$

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

$$8+x\times 2=x^{2}$$

Subtract x^{2} from both sides.

$$8+x\times 2-x^{2}=0$$

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.

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

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

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

Square 2.

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

Multiply -4 times -1.

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

Multiply 4 times 8.

$$x=\frac{-2±\sqrt{4+32}}{2\left(-1\right)}$$

Add 4 to 32.

$$x=\frac{-2±\sqrt{36}}{2\left(-1\right)}$$

Take the square root of 36.

$$x=\frac{-2±6}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-2±6}{-2}$$

Now solve the equation x=\frac{-2±6}{-2} when ± is plus. Add -2 to 6.

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

Divide 4 by -2.

$$x=-2$$

Now solve the equation x=\frac{-2±6}{-2} when ± is minus. Subtract 6 from -2.

$$x=-\frac{8}{-2}$$

Divide -8 by -2.

$$x=4$$

The equation is now solved.

$$x=-2 x=4$$

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.

$$8+x\times 2=xx$$

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

$$8+x\times 2=x^{2}$$

Subtract x^{2} from both sides.

$$8+x\times 2-x^{2}=0$$

Subtract 8 from both sides. Anything subtracted from zero gives its negation.

$$x\times 2-x^{2}=-8$$

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.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 2 by -1.

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

Divide -8 by -1.

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

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-2x+1=8+1$$

Add 8 to 1.

$$x^{2}-2x+1=9$$

Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-1\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{9}$$

Simplify.

$$x-1=3 x-1=-3$$

Add 1 to both sides of the equation.

$$x=4 x=-2$$