Solve 1/x+4+4/x-4-x^2-20/x^2-16=0

$$\frac{ 1 }{ x+4 } + \frac{ 4 }{ x-4 } - \frac{ { x }^{ 2 } -20 }{ { x }^{ 2 } -16 } = 0$$

Solve for x

$$x = \frac{3 \sqrt{17} + 5}{2} \approx 8.684658438$$

$$x=\frac{5-3\sqrt{17}}{2}\approx -3.684658438$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -4,4 since division by zero is not defined. Multiply both sides of the equation by \left(x-4\right)\left(x+4\right), the least common multiple of x+4,x-4,x^{2}-16.

$$x-4+\left(x+4\right)\times 4-\left(x^{2}-20\right)=0$$

Use the distributive property to multiply x+4 by 4.

$$x-4+4x+16-\left(x^{2}-20\right)=0$$

Combine x and 4x to get 5x.

$$5x-4+16-\left(x^{2}-20\right)=0$$

Add -4 and 16 to get 12.

$$5x+12-\left(x^{2}-20\right)=0$$

To find the opposite of x^{2}-20, find the opposite of each term.

$$5x+12-x^{2}+20=0$$

Add 12 and 20 to get 32.

$$5x+32-x^{2}=0$$

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.

$$-x^{2}+5x+32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 5 for b, and 32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-1\right)\times 32}}{2\left(-1\right)}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-1\right)\times 32}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-5±\sqrt{25+4\times 32}}{2\left(-1\right)}$$

Multiply 4 times 32.

$$x=\frac{-5±\sqrt{25+128}}{2\left(-1\right)}$$

Add 25 to 128.

$$x=\frac{-5±\sqrt{153}}{2\left(-1\right)}$$

Take the square root of 153.

$$x=\frac{-5±3\sqrt{17}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-5±3\sqrt{17}}{-2}$$

Now solve the equation x=\frac{-5±3\sqrt{17}}{-2} when ± is plus. Add -5 to 3\sqrt{17}.

$$x=\frac{3\sqrt{17}-5}{-2}$$

Divide -5+3\sqrt{17} by -2.

$$x=\frac{5-3\sqrt{17}}{2}$$

Now solve the equation x=\frac{-5±3\sqrt{17}}{-2} when ± is minus. Subtract 3\sqrt{17} from -5.

$$x=\frac{-3\sqrt{17}-5}{-2}$$

Divide -5-3\sqrt{17} by -2.

$$x=\frac{3\sqrt{17}+5}{2}$$

The equation is now solved.

$$x=\frac{5-3\sqrt{17}}{2} x=\frac{3\sqrt{17}+5}{2}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -4,4 since division by zero is not defined. Multiply both sides of the equation by \left(x-4\right)\left(x+4\right), the least common multiple of x+4,x-4,x^{2}-16.

$$x-4+\left(x+4\right)\times 4-\left(x^{2}-20\right)=0$$

Use the distributive property to multiply x+4 by 4.

$$x-4+4x+16-\left(x^{2}-20\right)=0$$

Combine x and 4x to get 5x.

$$5x-4+16-\left(x^{2}-20\right)=0$$

Add -4 and 16 to get 12.

$$5x+12-\left(x^{2}-20\right)=0$$

To find the opposite of x^{2}-20, find the opposite of each term.

$$5x+12-x^{2}+20=0$$

Add 12 and 20 to get 32.

$$5x+32-x^{2}=0$$

Subtract 32 from both sides. Anything subtracted from zero gives its negation.

$$5x-x^{2}=-32$$

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.

$$-x^{2}+5x=-32$$

Divide both sides by -1.

$$\frac{-x^{2}+5x}{-1}=-\frac{32}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{5}{-1}x=-\frac{32}{-1}$$

Divide 5 by -1.

$$x^{2}-5x=-\frac{32}{-1}$$

Divide -32 by -1.

$$x^{2}-5x=32$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=32+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=32+\frac{25}{4}$$

Add 32 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{153}{4}$$

Factor x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{153}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{153}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{3\sqrt{17}}{2} x-\frac{5}{2}=-\frac{3\sqrt{17}}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=\frac{3\sqrt{17}+5}{2} x=\frac{5-3\sqrt{17}}{2}$$