Solve 3/x+1+1/x-1=1

$$\frac { 3 } { x + 1 } + \frac { 1 } { x - 1 } = 1$$

Solve for x

$$x=\sqrt{3}+2\approx 3.732050808$$

$$x=2-\sqrt{3}\approx 0.267949192$$

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.

$$\left(x-1\right)\times 3+x+1=\left(x-1\right)\left(x+1\right)$$

Use the distributive property to multiply x-1 by 3.

$$3x-3+x+1=\left(x-1\right)\left(x+1\right)$$

Combine 3x and x to get 4x.

$$4x-3+1=\left(x-1\right)\left(x+1\right)$$

Add -3 and 1 to get -2.

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

Consider \left(x-1\right)\left(x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

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

Subtract x^{2} from both sides.

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

Add 1 to both sides.

$$4x-2-x^{2}+1=0$$

Add -2 and 1 to get -1.

$$4x-1-x^{2}=0$$

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.

$$-x^{2}+4x-1=0$$

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

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

Square 4.

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

Multiply -4 times -1.

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

Multiply 4 times -1.

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

Add 16 to -4.

$$x=\frac{-4±\sqrt{12}}{2\left(-1\right)}$$

Take the square root of 12.

$$x=\frac{-4±2\sqrt{3}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-4±2\sqrt{3}}{-2}$$

Now solve the equation x=\frac{-4±2\sqrt{3}}{-2} when ± is plus. Add -4 to 2\sqrt{3}.

$$x=\frac{2\sqrt{3}-4}{-2}$$

Divide -4+2\sqrt{3} by -2.

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

Now solve the equation x=\frac{-4±2\sqrt{3}}{-2} when ± is minus. Subtract 2\sqrt{3} from -4.

$$x=\frac{-2\sqrt{3}-4}{-2}$$

Divide -4-2\sqrt{3} by -2.

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

The equation is now solved.

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

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.

$$\left(x-1\right)\times 3+x+1=\left(x-1\right)\left(x+1\right)$$

Use the distributive property to multiply x-1 by 3.

$$3x-3+x+1=\left(x-1\right)\left(x+1\right)$$

Combine 3x and x to get 4x.

$$4x-3+1=\left(x-1\right)\left(x+1\right)$$

Add -3 and 1 to get -2.

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

Consider \left(x-1\right)\left(x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

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

Subtract x^{2} from both sides.

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

Add 2 to both sides.

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

Add -1 and 2 to get 1.

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

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.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 4 by -1.

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

Divide 1 by -1.

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

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

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

Square -2.

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

Add -1 to 4.

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

Factor x^{2}-4x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 2 to both sides of the equation.

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