Solve 3/x^2-9=1/x-3+x/2x+6

$$\frac{ 3 }{ { x }^{ 2 } -9 } = \frac{ 1 }{ x-3 } + \frac{ x }{ 2x+6 }$$

Solve for x

$$x=1$$

$$x=0$$

Steps Using Factoring

Variable x cannot be equal to any of the values -3,3 since division by zero is not defined. Multiply both sides of the equation by 2\left(x-3\right)\left(x+3\right), the least common multiple of x^{2}-9,x-3,2x+6.

$$2\times 3=2x+6+\left(x-3\right)x$$

Multiply 2 and 3 to get 6.

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

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

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

Combine 2x and -3x to get -x.

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

Swap sides so that all variable terms are on the left hand side.

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

Subtract 6 from both sides.

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

Subtract 6 from 6 to get 0.

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

Factor out x.

$$x\left(-1+x\right)=0$$

To find equation solutions, solve x=0 and -1+x=0.

$$x=0 x=1$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -3,3 since division by zero is not defined. Multiply both sides of the equation by 2\left(x-3\right)\left(x+3\right), the least common multiple of x^{2}-9,x-3,2x+6.

$$2\times 3=2x+6+\left(x-3\right)x$$

Multiply 2 and 3 to get 6.

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

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

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

Combine 2x and -3x to get -x.

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

Swap sides so that all variable terms are on the left hand side.

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

Subtract 6 from both sides.

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

Subtract 6 from 6 to get 0.

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

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.

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

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

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

Take the square root of 1.

$$x=\frac{-\left(-1\right)±1}{2}$$

The opposite of -1 is 1.

$$x=\frac{1±1}{2}$$

Now solve the equation x=\frac{1±1}{2} when ± is plus. Add 1 to 1.

$$x=\frac{2}{2}$$

Divide 2 by 2.

$$x=1$$

Now solve the equation x=\frac{1±1}{2} when ± is minus. Subtract 1 from 1.

$$x=\frac{0}{2}$$

Divide 0 by 2.

$$x=0$$

The equation is now solved.

$$x=1 x=0$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -3,3 since division by zero is not defined. Multiply both sides of the equation by 2\left(x-3\right)\left(x+3\right), the least common multiple of x^{2}-9,x-3,2x+6.

$$2\times 3=2x+6+\left(x-3\right)x$$

Multiply 2 and 3 to get 6.

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

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

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

Combine 2x and -3x to get -x.

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

Swap sides so that all variable terms are on the left hand side.

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

Subtract 6 from both sides.

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

Subtract 6 from 6 to get 0.

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

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.

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

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

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Factor x^{2}-x+\frac{1}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{1}{2} to both sides of the equation.

$$x=1 x=0$$