Solve 2+x=1/x-5

$$2 + x = \frac { 1 } { x - 5 }$$

Solve for x

$$x = \frac{\sqrt{53} + 3}{2} \approx 5.140054945$$

$$x=\frac{3-\sqrt{53}}{2}\approx -2.140054945$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 5 since division by zero is not defined. Multiply both sides of the equation by x-5.

$$\left(x-5\right)\times 2+\left(x-5\right)x=1$$

Use the distributive property to multiply x-5 by 2.

$$2x-10+\left(x-5\right)x=1$$

Use the distributive property to multiply x-5 by x.

$$2x-10+x^{2}-5x=1$$

Combine 2x and -5x to get -3x.

$$-3x-10+x^{2}=1$$

Subtract 1 from both sides.

$$-3x-10+x^{2}-1=0$$

Subtract 1 from -10 to get -11.

$$-3x-11+x^{2}=0$$

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.

$$x^{2}-3x-11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and -11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-11\right)}}{2}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-11\right)}}{2}$$

Multiply -4 times -11.

$$x=\frac{-\left(-3\right)±\sqrt{9+44}}{2}$$

Add 9 to 44.

$$x=\frac{-\left(-3\right)±\sqrt{53}}{2}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{53}}{2}$$

Now solve the equation x=\frac{3±\sqrt{53}}{2} when ± is plus. Add 3 to \sqrt{53}.

$$x=\frac{\sqrt{53}+3}{2}$$

Now solve the equation x=\frac{3±\sqrt{53}}{2} when ± is minus. Subtract \sqrt{53} from 3.

$$x=\frac{3-\sqrt{53}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{53}+3}{2} x=\frac{3-\sqrt{53}}{2}$$

Steps for Completing the Square

Variable x cannot be equal to 5 since division by zero is not defined. Multiply both sides of the equation by x-5.

$$\left(x-5\right)\times 2+\left(x-5\right)x=1$$

Use the distributive property to multiply x-5 by 2.

$$2x-10+\left(x-5\right)x=1$$

Use the distributive property to multiply x-5 by x.

$$2x-10+x^{2}-5x=1$$

Combine 2x and -5x to get -3x.

$$-3x-10+x^{2}=1$$

Add 10 to both sides.

$$-3x+x^{2}=1+10$$

Add 1 and 10 to get 11.

$$-3x+x^{2}=11$$

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.

$$x^{2}-3x=11$$

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

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=11+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=11+\frac{9}{4}$$

Add 11 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{53}{4}$$

Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{53}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{53}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{53}}{2} x-\frac{3}{2}=-\frac{\sqrt{53}}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{\sqrt{53}+3}{2} x=\frac{3-\sqrt{53}}{2}$$