Solve 3/x=x-2

$$\frac{ 3 }{ x } =x-2$$

Solve for x

$$x=-1$$

$$x=3$$

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.

$$3=xx+x\left(-2\right)$$

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

$$3=x^{2}+x\left(-2\right)$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}+x\left(-2\right)=3$$

Subtract 3 from both sides.

$$x^{2}+x\left(-2\right)-3=0$$

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.

$$x^{2}-2x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -2 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-3\right)}}{2}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-3\right)}}{2}$$

Multiply -4 times -3.

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

Add 4 to 12.

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

Take the square root of 16.

$$x=\frac{-\left(-2\right)±4}{2}$$

The opposite of -2 is 2.

$$x=\frac{2±4}{2}$$

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

$$x=\frac{6}{2}$$

Divide 6 by 2.

$$x=3$$

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

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

Divide -2 by 2.

$$x=-1$$

The equation is now solved.

$$x=3 x=-1$$

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.

$$3=xx+x\left(-2\right)$$

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

$$3=x^{2}+x\left(-2\right)$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}+x\left(-2\right)=3$$

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.

$$x^{2}-2x=3$$

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.

$$x^{2}-2x+1=3+1$$

Add 3 to 1.

$$x^{2}-2x+1=4$$

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}.

$$\left(x-1\right)^{2}=4$$

Take the square root of both sides of the equation.

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

Simplify.

$$x-1=2 x-1=-2$$

Add 1 to both sides of the equation.

$$x=3 x=-1$$