Solve 2x+1=x+1/x-5

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

Solve for x

$$x = \frac{\sqrt{37} + 5}{2} \approx 5.541381265$$

$$x=\frac{5-\sqrt{37}}{2}\approx -0.541381265$$

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.

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

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

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

Combine -10x and x to get -9x.

$$2x^{2}-9x-5=x+1$$

Subtract x from both sides.

$$2x^{2}-9x-5-x=1$$

Combine -9x and -x to get -10x.

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

Subtract 1 from both sides.

$$2x^{2}-10x-5-1=0$$

Subtract 1 from -5 to get -6.

$$2x^{2}-10x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -10 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 2\left(-6\right)}}{2\times 2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 2\left(-6\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-10\right)±\sqrt{100-8\left(-6\right)}}{2\times 2}$$

Multiply -8 times -6.

$$x=\frac{-\left(-10\right)±\sqrt{100+48}}{2\times 2}$$

Add 100 to 48.

$$x=\frac{-\left(-10\right)±\sqrt{148}}{2\times 2}$$

Take the square root of 148.

$$x=\frac{-\left(-10\right)±2\sqrt{37}}{2\times 2}$$

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{37}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{10±2\sqrt{37}}{4}$$

Now solve the equation x=\frac{10±2\sqrt{37}}{4} when ± is plus. Add 10 to 2\sqrt{37}.

$$x=\frac{2\sqrt{37}+10}{4}$$

Divide 10+2\sqrt{37} by 4.

$$x=\frac{\sqrt{37}+5}{2}$$

Now solve the equation x=\frac{10±2\sqrt{37}}{4} when ± is minus. Subtract 2\sqrt{37} from 10.

$$x=\frac{10-2\sqrt{37}}{4}$$

Divide 10-2\sqrt{37} by 4.

$$x=\frac{5-\sqrt{37}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{37}+5}{2} x=\frac{5-\sqrt{37}}{2}$$

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.

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

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

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

Combine -10x and x to get -9x.

$$2x^{2}-9x-5=x+1$$

Subtract x from both sides.

$$2x^{2}-9x-5-x=1$$

Combine -9x and -x to get -10x.

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

Add 5 to both sides.

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

Add 1 and 5 to get 6.

$$2x^{2}-10x=6$$

Divide both sides by 2.

$$\frac{2x^{2}-10x}{2}=\frac{6}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{10}{2}\right)x=\frac{6}{2}$$

Divide -10 by 2.

$$x^{2}-5x=\frac{6}{2}$$

Divide 6 by 2.

$$x^{2}-5x=3$$

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

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=3+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=3+\frac{25}{4}$$

Add 3 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{37}{4}$$

Factor x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{37}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{37}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{\sqrt{37}}{2} x-\frac{5}{2}=-\frac{\sqrt{37}}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=\frac{\sqrt{37}+5}{2} x=\frac{5-\sqrt{37}}{2}$$