Solve 1/x-1+1/x+1=4

$$\frac { 1 } { x - 1 } + \frac { 1 } { x + 1 } = 4$$

Solve for x

$$x = \frac{\sqrt{17} + 1}{4} \approx 1.280776406$$

$$x=\frac{1-\sqrt{17}}{4}\approx -0.780776406$$

Steps Using the Quadratic Formula

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

$$x+1+x-1=4\left(x-1\right)\left(x+1\right)$$

Combine x and x to get 2x.

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

Subtract 1 from 1 to get 0.

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

Use the distributive property to multiply 4 by x-1.

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

Use the distributive property to multiply 4x-4 by x+1 and combine like terms.

$$2x=4x^{2}-4$$

Subtract 4x^{2} from both sides.

$$2x-4x^{2}=-4$$

Add 4 to both sides.

$$2x-4x^{2}+4=0$$

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.

$$-4x^{2}+2x+4=0$$

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

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

Square 2.

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

Multiply -4 times -4.

$$x=\frac{-2±\sqrt{4+16\times 4}}{2\left(-4\right)}$$

Multiply 16 times 4.

$$x=\frac{-2±\sqrt{4+64}}{2\left(-4\right)}$$

Add 4 to 64.

$$x=\frac{-2±\sqrt{68}}{2\left(-4\right)}$$

Take the square root of 68.

$$x=\frac{-2±2\sqrt{17}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-2±2\sqrt{17}}{-8}$$

Now solve the equation x=\frac{-2±2\sqrt{17}}{-8} when ± is plus. Add -2 to 2\sqrt{17}.

$$x=\frac{2\sqrt{17}-2}{-8}$$

Divide -2+2\sqrt{17} by -8.

$$x=\frac{1-\sqrt{17}}{4}$$

Now solve the equation x=\frac{-2±2\sqrt{17}}{-8} when ± is minus. Subtract 2\sqrt{17} from -2.

$$x=\frac{-2\sqrt{17}-2}{-8}$$

Divide -2-2\sqrt{17} by -8.

$$x=\frac{\sqrt{17}+1}{4}$$

The equation is now solved.

$$x=\frac{1-\sqrt{17}}{4} x=\frac{\sqrt{17}+1}{4}$$

Steps for Completing the Square

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

$$x+1+x-1=4\left(x-1\right)\left(x+1\right)$$

Combine x and x to get 2x.

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

Subtract 1 from 1 to get 0.

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

Use the distributive property to multiply 4 by x-1.

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

Use the distributive property to multiply 4x-4 by x+1 and combine like terms.

$$2x=4x^{2}-4$$

Subtract 4x^{2} from both sides.

$$2x-4x^{2}=-4$$

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.

$$-4x^{2}+2x=-4$$

Divide both sides by -4.

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

Dividing by -4 undoes the multiplication by -4.

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

Reduce the fraction \frac{2}{-4} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{1}{2}x=-\frac{4}{-4}$$

Divide -4 by -4.

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

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=1+\frac{1}{16}$$

Add 1 to \frac{1}{16}.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{17}{16}$$

Factor x^{2}-\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{17}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{17}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{\sqrt{17}}{4} x-\frac{1}{4}=-\frac{\sqrt{17}}{4}$$

Add \frac{1}{4} to both sides of the equation.

$$x=\frac{\sqrt{17}+1}{4} x=\frac{1-\sqrt{17}}{4}$$