Solve x^2/x-2-6x-8/x-2=/x-2

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

Solve for x

$$x=\sqrt{2}+3\approx 4.414213562$$

$$x=3-\sqrt{2}\approx 1.585786438$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2.

$$x^{2}-\left(6x-8\right)=1$$

To find the opposite of 6x-8, find the opposite of each term.

$$x^{2}-6x+8=1$$

Subtract 1 from both sides.

$$x^{2}-6x+8-1=0$$

Subtract 1 from 8 to get 7.

$$x^{2}-6x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -6 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 7}}{2}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 7}}{2}$$

Multiply -4 times 7.

$$x=\frac{-\left(-6\right)±\sqrt{36-28}}{2}$$

Add 36 to -28.

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

Take the square root of 8.

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

The opposite of -6 is 6.

$$x=\frac{6±2\sqrt{2}}{2}$$

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

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

Divide 6+2\sqrt{2} by 2.

$$x=\sqrt{2}+3$$

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

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

Divide 6-2\sqrt{2} by 2.

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

The equation is now solved.

$$x=\sqrt{2}+3 x=3-\sqrt{2}$$

Steps for Completing the Square

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2.

$$x^{2}-\left(6x-8\right)=1$$

To find the opposite of 6x-8, find the opposite of each term.

$$x^{2}-6x+8=1$$

Subtract 8 from both sides.

$$x^{2}-6x=1-8$$

Subtract 8 from 1 to get -7.

$$x^{2}-6x=-7$$

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.

$$x^{2}-6x+\left(-3\right)^{2}=-7+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-7+9$$

Add -7 to 9.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 3 to both sides of the equation.

$$x=\sqrt{2}+3 x=3-\sqrt{2}$$