Solve 1/x+3+1/x-3=1/4

$$\frac{ 1 }{ x+3 } + \frac{ 1 }{ x-3 } = \frac{ 1 }{ 4 }$$

Solve for x

$$x=-1$$

$$x=9$$

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 4\left(x-3\right)\left(x+3\right), the least common multiple of x+3,x-3,4.

$$4x-12+4x+12=\left(x-3\right)\left(x+3\right)$$

Combine 4x and 4x to get 8x.

$$8x-12+12=\left(x-3\right)\left(x+3\right)$$

Add -12 and 12 to get 0.

$$8x=\left(x-3\right)\left(x+3\right)$$

Consider \left(x-3\right)\left(x+3\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 3.

$$8x=x^{2}-9$$

Subtract x^{2} from both sides.

$$8x-x^{2}=-9$$

Add 9 to both sides.

$$8x-x^{2}+9=0$$

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.

$$-x^{2}+8x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 8 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-1\right)\times 9}}{2\left(-1\right)}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-1\right)\times 9}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-8±\sqrt{64+4\times 9}}{2\left(-1\right)}$$

Multiply 4 times 9.

$$x=\frac{-8±\sqrt{64+36}}{2\left(-1\right)}$$

Add 64 to 36.

$$x=\frac{-8±\sqrt{100}}{2\left(-1\right)}$$

Take the square root of 100.

$$x=\frac{-8±10}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-8±10}{-2}$$

Now solve the equation x=\frac{-8±10}{-2} when ± is plus. Add -8 to 10.

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

Divide 2 by -2.

$$x=-1$$

Now solve the equation x=\frac{-8±10}{-2} when ± is minus. Subtract 10 from -8.

$$x=-\frac{18}{-2}$$

Divide -18 by -2.

$$x=9$$

The equation is now solved.

$$x=-1 x=9$$

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 4\left(x-3\right)\left(x+3\right), the least common multiple of x+3,x-3,4.

$$4x-12+4x+12=\left(x-3\right)\left(x+3\right)$$

Combine 4x and 4x to get 8x.

$$8x-12+12=\left(x-3\right)\left(x+3\right)$$

Add -12 and 12 to get 0.

$$8x=\left(x-3\right)\left(x+3\right)$$

Consider \left(x-3\right)\left(x+3\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 3.

$$8x=x^{2}-9$$

Subtract x^{2} from both sides.

$$8x-x^{2}=-9$$

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.

$$-x^{2}+8x=-9$$

Divide both sides by -1.

$$\frac{-x^{2}+8x}{-1}=-\frac{9}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{8}{-1}x=-\frac{9}{-1}$$

Divide 8 by -1.

$$x^{2}-8x=-\frac{9}{-1}$$

Divide -9 by -1.

$$x^{2}-8x=9$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=9+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=9+16$$

Add 9 to 16.

$$x^{2}-8x+16=25$$

Factor x^{2}-8x+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}.

$$\left(x-4\right)^{2}=25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{25}$$

Simplify.

$$x-4=5 x-4=-5$$

Add 4 to both sides of the equation.

$$x=9 x=-1$$