Solve x+1/x+5-1/5-x=4

$$\frac { x + 1 } { x + 5 } - \frac { 1 } { 5 - x } = 4$$

Solve for x

$$x=\frac{\sqrt{1209}}{6}-\frac{1}{2}\approx 5.295112884$$

$$x=-\frac{\sqrt{1209}}{6}-\frac{1}{2}\approx -6.295112884$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -5,5 since division by zero is not defined. Multiply both sides of the equation by \left(x-5\right)\left(x+5\right), the least common multiple of x+5,5-x.

$$\left(x-5\right)\left(x+1\right)+5+x=4\left(x-5\right)\left(x+5\right)$$

Use the distributive property to multiply x-5 by x+1 and combine like terms.

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

Add -5 and 5 to get 0.

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

Combine -4x and x to get -3x.

$$x^{2}-3x=4\left(x-5\right)\left(x+5\right)$$

Use the distributive property to multiply 4 by x-5.

$$x^{2}-3x=\left(4x-20\right)\left(x+5\right)$$

Use the distributive property to multiply 4x-20 by x+5 and combine like terms.

$$x^{2}-3x=4x^{2}-100$$

Subtract 4x^{2} from both sides.

$$x^{2}-3x-4x^{2}=-100$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}-3x=-100$$

Add 100 to both sides.

$$-3x^{2}-3x+100=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -3 for b, and 100 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-3\right)\times 100}}{2\left(-3\right)}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-3\right)\times 100}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-3\right)±\sqrt{9+12\times 100}}{2\left(-3\right)}$$

Multiply 12 times 100.

$$x=\frac{-\left(-3\right)±\sqrt{9+1200}}{2\left(-3\right)}$$

Add 9 to 1200.

$$x=\frac{-\left(-3\right)±\sqrt{1209}}{2\left(-3\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{1209}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{3±\sqrt{1209}}{-6}$$

Now solve the equation x=\frac{3±\sqrt{1209}}{-6} when ± is plus. Add 3 to \sqrt{1209}.

$$x=\frac{\sqrt{1209}+3}{-6}$$

Divide 3+\sqrt{1209} by -6.

$$x=-\frac{\sqrt{1209}}{6}-\frac{1}{2}$$

Now solve the equation x=\frac{3±\sqrt{1209}}{-6} when ± is minus. Subtract \sqrt{1209} from 3.

$$x=\frac{3-\sqrt{1209}}{-6}$$

Divide 3-\sqrt{1209} by -6.

$$x=\frac{\sqrt{1209}}{6}-\frac{1}{2}$$

The equation is now solved.

$$x=-\frac{\sqrt{1209}}{6}-\frac{1}{2} x=\frac{\sqrt{1209}}{6}-\frac{1}{2}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -5,5 since division by zero is not defined. Multiply both sides of the equation by \left(x-5\right)\left(x+5\right), the least common multiple of x+5,5-x.

$$\left(x-5\right)\left(x+1\right)+5+x=4\left(x-5\right)\left(x+5\right)$$

Use the distributive property to multiply x-5 by x+1 and combine like terms.

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

Add -5 and 5 to get 0.

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

Combine -4x and x to get -3x.

$$x^{2}-3x=4\left(x-5\right)\left(x+5\right)$$

Use the distributive property to multiply 4 by x-5.

$$x^{2}-3x=\left(4x-20\right)\left(x+5\right)$$

Use the distributive property to multiply 4x-20 by x+5 and combine like terms.

$$x^{2}-3x=4x^{2}-100$$

Subtract 4x^{2} from both sides.

$$x^{2}-3x-4x^{2}=-100$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}-3x=-100$$

Divide both sides by -3.

$$\frac{-3x^{2}-3x}{-3}=-\frac{100}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\left(-\frac{3}{-3}\right)x=-\frac{100}{-3}$$

Divide -3 by -3.

$$x^{2}+x=-\frac{100}{-3}$$

Divide -100 by -3.

$$x^{2}+x=\frac{100}{3}$$

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.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=\frac{100}{3}+\left(\frac{1}{2}\right)^{2}$$

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

$$x^{2}+x+\frac{1}{4}=\frac{100}{3}+\frac{1}{4}$$

Add \frac{100}{3} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+x+\frac{1}{4}=\frac{403}{12}$$

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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{403}{12}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{403}{12}}$$

Simplify.

$$x+\frac{1}{2}=\frac{\sqrt{1209}}{6} x+\frac{1}{2}=-\frac{\sqrt{1209}}{6}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=\frac{\sqrt{1209}}{6}-\frac{1}{2} x=-\frac{\sqrt{1209}}{6}-\frac{1}{2}$$