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

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

Solve for x

$$x=\frac{\sqrt{13}-5}{2}\approx -0.697224362$$

$$x=\frac{-\sqrt{13}-5}{2}\approx -4.302775638$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -\frac{1}{2},1 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)\left(2x+1\right), the least common multiple of 2x+1,x-1.

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

Multiply 2x+1 and 2x+1 to get \left(2x+1\right)^{2}.

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

Use the distributive property to multiply x-1 by x.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

$$x^{2}-x+4x^{2}+4x+1=2\left(x-1\right)\left(2x+1\right)$$

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$5x^{2}-x+4x+1=2\left(x-1\right)\left(2x+1\right)$$

Combine -x and 4x to get 3x.

$$5x^{2}+3x+1=2\left(x-1\right)\left(2x+1\right)$$

Use the distributive property to multiply 2 by x-1.

$$5x^{2}+3x+1=\left(2x-2\right)\left(2x+1\right)$$

Use the distributive property to multiply 2x-2 by 2x+1 and combine like terms.

$$5x^{2}+3x+1=4x^{2}-2x-2$$

Subtract 4x^{2} from both sides.

$$5x^{2}+3x+1-4x^{2}=-2x-2$$

Combine 5x^{2} and -4x^{2} to get x^{2}.

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

Add 2x to both sides.

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

Combine 3x and 2x to get 5x.

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

Add 2 to both sides.

$$x^{2}+5x+1+2=0$$

Add 1 and 2 to get 3.

$$x^{2}+5x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 5 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\times 3}}{2}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 3}}{2}$$

Multiply -4 times 3.

$$x=\frac{-5±\sqrt{25-12}}{2}$$

Add 25 to -12.

$$x=\frac{-5±\sqrt{13}}{2}$$

Now solve the equation x=\frac{-5±\sqrt{13}}{2} when ± is plus. Add -5 to \sqrt{13}.

$$x=\frac{\sqrt{13}-5}{2}$$

Now solve the equation x=\frac{-5±\sqrt{13}}{2} when ± is minus. Subtract \sqrt{13} from -5.

$$x=\frac{-\sqrt{13}-5}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{13}-5}{2} x=\frac{-\sqrt{13}-5}{2}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -\frac{1}{2},1 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)\left(2x+1\right), the least common multiple of 2x+1,x-1.

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

Multiply 2x+1 and 2x+1 to get \left(2x+1\right)^{2}.

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

Use the distributive property to multiply x-1 by x.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

$$x^{2}-x+4x^{2}+4x+1=2\left(x-1\right)\left(2x+1\right)$$

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$5x^{2}-x+4x+1=2\left(x-1\right)\left(2x+1\right)$$

Combine -x and 4x to get 3x.

$$5x^{2}+3x+1=2\left(x-1\right)\left(2x+1\right)$$

Use the distributive property to multiply 2 by x-1.

$$5x^{2}+3x+1=\left(2x-2\right)\left(2x+1\right)$$

Use the distributive property to multiply 2x-2 by 2x+1 and combine like terms.

$$5x^{2}+3x+1=4x^{2}-2x-2$$

Subtract 4x^{2} from both sides.

$$5x^{2}+3x+1-4x^{2}=-2x-2$$

Combine 5x^{2} and -4x^{2} to get x^{2}.

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

Add 2x to both sides.

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

Combine 3x and 2x to get 5x.

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

Subtract 1 from both sides.

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

Subtract 1 from -2 to get -3.

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

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}.

$$x^{2}+5x+\frac{25}{4}=\frac{13}{4}$$

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}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{13}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{13}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{\sqrt{13}}{2} x+\frac{5}{2}=-\frac{\sqrt{13}}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=\frac{\sqrt{13}-5}{2} x=\frac{-\sqrt{13}-5}{2}$$