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

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

Solve for x

$$x=\sqrt{7}+1\approx 3.645751311$$

$$x=1-\sqrt{7}\approx -1.645751311$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -3,0,3 since division by zero is not defined. Multiply both sides of the equation by x\left(x-3\right)\left(x+3\right), the least common multiple of x^{2}+3x,x^{2}-9,x.

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

Combine x and x to get 2x.

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

Subtract x^{2} from both sides.

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

Add 9 to both sides.

$$2x-3-x^{2}+9=0$$

Add -3 and 9 to get 6.

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

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.

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

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

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

Square 2.

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

Multiply -4 times -1.

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

Multiply 4 times 6.

$$x=\frac{-2±\sqrt{4+24}}{2\left(-1\right)}$$

Add 4 to 24.

$$x=\frac{-2±\sqrt{28}}{2\left(-1\right)}$$

Take the square root of 28.

$$x=\frac{-2±2\sqrt{7}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-2±2\sqrt{7}}{-2}$$

Now solve the equation x=\frac{-2±2\sqrt{7}}{-2} when ± is plus. Add -2 to 2\sqrt{7}.

$$x=\frac{2\sqrt{7}-2}{-2}$$

Divide -2+2\sqrt{7} by -2.

$$x=1-\sqrt{7}$$

Now solve the equation x=\frac{-2±2\sqrt{7}}{-2} when ± is minus. Subtract 2\sqrt{7} from -2.

$$x=\frac{-2\sqrt{7}-2}{-2}$$

Divide -2-2\sqrt{7} by -2.

$$x=\sqrt{7}+1$$

The equation is now solved.

$$x=1-\sqrt{7} x=\sqrt{7}+1$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -3,0,3 since division by zero is not defined. Multiply both sides of the equation by x\left(x-3\right)\left(x+3\right), the least common multiple of x^{2}+3x,x^{2}-9,x.

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

Combine x and x to get 2x.

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

Subtract x^{2} from both sides.

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

Add 3 to both sides.

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

Add -9 and 3 to get -6.

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

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.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 2 by -1.

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

Divide -6 by -1.

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

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

$$x^{2}-2x+1=6+1$$

Add 6 to 1.

$$x^{2}-2x+1=7$$

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

$$\left(x-1\right)^{2}=7$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{7}$$

Simplify.

$$x-1=\sqrt{7} x-1=-\sqrt{7}$$

Add 1 to both sides of the equation.

$$x=\sqrt{7}+1 x=1-\sqrt{7}$$