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

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

Solve for x

$$x=2\sqrt{3}+3\approx 6.464101615$$

$$x=3-2\sqrt{3}\approx -0.464101615$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -1,0,3 since division by zero is not defined. Multiply both sides of the equation by x\left(x-3\right)\left(x+1\right), the least common multiple of x-3,x+1,x.

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

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

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

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

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

To find the opposite of x^{2}-3x, find the opposite of each term.

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

Combine x^{2} and -x^{2} to get 0.

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

Combine x and 3x to get 4x.

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

Use the distributive property to multiply x-3 by x+1 and combine like terms.

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

Subtract x^{2} from both sides.

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

Add 2x to both sides.

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

Combine 4x and 2x to get 6x.

$$6x-x^{2}=-3$$

Add 3 to both sides.

$$6x-x^{2}+3=0$$

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.

$$-x^{2}+6x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 6 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-1\right)\times 3}}{2\left(-1\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-1\right)\times 3}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-6±\sqrt{36+4\times 3}}{2\left(-1\right)}$$

Multiply 4 times 3.

$$x=\frac{-6±\sqrt{36+12}}{2\left(-1\right)}$$

Add 36 to 12.

$$x=\frac{-6±\sqrt{48}}{2\left(-1\right)}$$

Take the square root of 48.

$$x=\frac{-6±4\sqrt{3}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-6±4\sqrt{3}}{-2}$$

Now solve the equation x=\frac{-6±4\sqrt{3}}{-2} when ± is plus. Add -6 to 4\sqrt{3}.

$$x=\frac{4\sqrt{3}-6}{-2}$$

Divide -6+4\sqrt{3} by -2.

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

Now solve the equation x=\frac{-6±4\sqrt{3}}{-2} when ± is minus. Subtract 4\sqrt{3} from -6.

$$x=\frac{-4\sqrt{3}-6}{-2}$$

Divide -6-4\sqrt{3} by -2.

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

The equation is now solved.

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

Steps for Completing the Square

Variable x cannot be equal to any of the values -1,0,3 since division by zero is not defined. Multiply both sides of the equation by x\left(x-3\right)\left(x+1\right), the least common multiple of x-3,x+1,x.

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

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

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

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

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

To find the opposite of x^{2}-3x, find the opposite of each term.

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

Combine x^{2} and -x^{2} to get 0.

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

Combine x and 3x to get 4x.

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

Use the distributive property to multiply x-3 by x+1 and combine like terms.

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

Subtract x^{2} from both sides.

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

Add 2x to both sides.

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

Combine 4x and 2x to get 6x.

$$6x-x^{2}=-3$$

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.

$$-x^{2}+6x=-3$$

Divide both sides by -1.

$$\frac{-x^{2}+6x}{-1}=-\frac{3}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{6}{-1}x=-\frac{3}{-1}$$

Divide 6 by -1.

$$x^{2}-6x=-\frac{3}{-1}$$

Divide -3 by -1.

$$x^{2}-6x=3$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=3+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=3+9$$

Add 3 to 9.

$$x^{2}-6x+9=12$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=12$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{12}$$

Simplify.

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

Add 3 to both sides of the equation.

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