Solve x-2=4x-8/x

$$x - 2 = \frac { 4 x - 8 } { x }$$

Solve for x

$$x=2$$

$$x=4$$

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 x.

$$xx+x\left(-2\right)=4x-8$$

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

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

Subtract 4x from both sides.

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

Combine x\left(-2\right) and -4x to get -6x.

$$x^{2}-6x=-8$$

Add 8 to both sides.

$$x^{2}-6x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -6 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 8}}{2}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 8}}{2}$$

Multiply -4 times 8.

$$x=\frac{-\left(-6\right)±\sqrt{36-32}}{2}$$

Add 36 to -32.

$$x=\frac{-\left(-6\right)±\sqrt{4}}{2}$$

Take the square root of 4.

$$x=\frac{-\left(-6\right)±2}{2}$$

The opposite of -6 is 6.

$$x=\frac{6±2}{2}$$

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

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

Divide 8 by 2.

$$x=4$$

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

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

Divide 4 by 2.

$$x=2$$

The equation is now solved.

$$x=4 x=2$$

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.

$$xx+x\left(-2\right)=4x-8$$

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

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

Subtract 4x from both sides.

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

Combine x\left(-2\right) and -4x to get -6x.

$$x^{2}-6x=-8$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=-8+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-8+9$$

Add -8 to 9.

$$x^{2}-6x+9=1$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{1}$$

Simplify.

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

Add 3 to both sides of the equation.

$$x=4 x=2$$