Solve (x)=x-2/x+11

$$( x ) = \frac { x - 2 } { x + 11 }$$

Solve for x (complex solution)

$$x=\sqrt{23}-5\approx -0.204168477$$

$$x=-\left(\sqrt{23}+5\right)\approx -9.795831523$$

Steps Using the Quadratic Formula

Subtract \frac{x-2}{x+11} from both sides.

$$x-\frac{x-2}{x+11}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{x+11}{x+11}.

$$\frac{x\left(x+11\right)}{x+11}-\frac{x-2}{x+11}=0$$

Since \frac{x\left(x+11\right)}{x+11} and \frac{x-2}{x+11} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(x+11\right)-\left(x-2\right)}{x+11}=0$$

Do the multiplications in x\left(x+11\right)-\left(x-2\right).

$$\frac{x^{2}+11x-x+2}{x+11}=0$$

Combine like terms in x^{2}+11x-x+2.

$$\frac{x^{2}+10x+2}{x+11}=0$$

Variable x cannot be equal to -11 since division by zero is not defined. Multiply both sides of the equation by x+11.

$$x^{2}+10x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 10 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\times 2}}{2}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 2}}{2}$$

Multiply -4 times 2.

$$x=\frac{-10±\sqrt{100-8}}{2}$$

Add 100 to -8.

$$x=\frac{-10±\sqrt{92}}{2}$$

Take the square root of 92.

$$x=\frac{-10±2\sqrt{23}}{2}$$

Now solve the equation x=\frac{-10±2\sqrt{23}}{2} when ± is plus. Add -10 to 2\sqrt{23}.

$$x=\frac{2\sqrt{23}-10}{2}$$

Divide -10+2\sqrt{23} by 2.

$$x=\sqrt{23}-5$$

Now solve the equation x=\frac{-10±2\sqrt{23}}{2} when ± is minus. Subtract 2\sqrt{23} from -10.

$$x=\frac{-2\sqrt{23}-10}{2}$$

Divide -10-2\sqrt{23} by 2.

$$x=-\sqrt{23}-5$$

The equation is now solved.

$$x=\sqrt{23}-5 x=-\sqrt{23}-5$$

Steps for Completing the Square

Subtract \frac{x-2}{x+11} from both sides.

$$x-\frac{x-2}{x+11}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{x+11}{x+11}.

$$\frac{x\left(x+11\right)}{x+11}-\frac{x-2}{x+11}=0$$

Since \frac{x\left(x+11\right)}{x+11} and \frac{x-2}{x+11} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(x+11\right)-\left(x-2\right)}{x+11}=0$$

Do the multiplications in x\left(x+11\right)-\left(x-2\right).

$$\frac{x^{2}+11x-x+2}{x+11}=0$$

Combine like terms in x^{2}+11x-x+2.

$$\frac{x^{2}+10x+2}{x+11}=0$$

Variable x cannot be equal to -11 since division by zero is not defined. Multiply both sides of the equation by x+11.

$$x^{2}+10x+2=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}+10x=-2$$

Divide 10, the coefficient of the x term, by 2 to get 5. Then add the square of 5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+10x+5^{2}=-2+5^{2}$$

Square 5.

$$x^{2}+10x+25=-2+25$$

Add -2 to 25.

$$x^{2}+10x+25=23$$

Factor x^{2}+10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+5\right)^{2}=23$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{23}$$

Simplify.

$$x+5=\sqrt{23} x+5=-\sqrt{23}$$

Subtract 5 from both sides of the equation.

$$x=\sqrt{23}-5 x=-\sqrt{23}-5$$

Solve for x

$$x=\sqrt{23}-5\approx -0.204168477$$

$$x=-\sqrt{23}-5\approx -9.795831523$$

Steps Using the Quadratic Formula

Subtract \frac{x-2}{x+11} from both sides.

$$x-\frac{x-2}{x+11}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{x+11}{x+11}.

$$\frac{x\left(x+11\right)}{x+11}-\frac{x-2}{x+11}=0$$

Since \frac{x\left(x+11\right)}{x+11} and \frac{x-2}{x+11} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(x+11\right)-\left(x-2\right)}{x+11}=0$$

Do the multiplications in x\left(x+11\right)-\left(x-2\right).

$$\frac{x^{2}+11x-x+2}{x+11}=0$$

Combine like terms in x^{2}+11x-x+2.

$$\frac{x^{2}+10x+2}{x+11}=0$$

Variable x cannot be equal to -11 since division by zero is not defined. Multiply both sides of the equation by x+11.

$$x^{2}+10x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 10 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\times 2}}{2}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 2}}{2}$$

Multiply -4 times 2.

$$x=\frac{-10±\sqrt{100-8}}{2}$$

Add 100 to -8.

$$x=\frac{-10±\sqrt{92}}{2}$$

Take the square root of 92.

$$x=\frac{-10±2\sqrt{23}}{2}$$

Now solve the equation x=\frac{-10±2\sqrt{23}}{2} when ± is plus. Add -10 to 2\sqrt{23}.

$$x=\frac{2\sqrt{23}-10}{2}$$

Divide -10+2\sqrt{23} by 2.

$$x=\sqrt{23}-5$$

Now solve the equation x=\frac{-10±2\sqrt{23}}{2} when ± is minus. Subtract 2\sqrt{23} from -10.

$$x=\frac{-2\sqrt{23}-10}{2}$$

Divide -10-2\sqrt{23} by 2.

$$x=-\sqrt{23}-5$$

The equation is now solved.

$$x=\sqrt{23}-5 x=-\sqrt{23}-5$$

Steps for Completing the Square

Subtract \frac{x-2}{x+11} from both sides.

$$x-\frac{x-2}{x+11}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{x+11}{x+11}.

$$\frac{x\left(x+11\right)}{x+11}-\frac{x-2}{x+11}=0$$

Since \frac{x\left(x+11\right)}{x+11} and \frac{x-2}{x+11} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(x+11\right)-\left(x-2\right)}{x+11}=0$$

Do the multiplications in x\left(x+11\right)-\left(x-2\right).

$$\frac{x^{2}+11x-x+2}{x+11}=0$$

Combine like terms in x^{2}+11x-x+2.

$$\frac{x^{2}+10x+2}{x+11}=0$$

Variable x cannot be equal to -11 since division by zero is not defined. Multiply both sides of the equation by x+11.

$$x^{2}+10x+2=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}+10x=-2$$

Divide 10, the coefficient of the x term, by 2 to get 5. Then add the square of 5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+10x+5^{2}=-2+5^{2}$$

Square 5.

$$x^{2}+10x+25=-2+25$$

Add -2 to 25.

$$x^{2}+10x+25=23$$

Factor x^{2}+10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+5\right)^{2}=23$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{23}$$

Simplify.

$$x+5=\sqrt{23} x+5=-\sqrt{23}$$

Subtract 5 from both sides of the equation.

$$x=\sqrt{23}-5 x=-\sqrt{23}-5$$