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

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

Solve for x

$$x=\frac{\sqrt{209}-17}{10}\approx -0.254316771$$

$$x=\frac{-\sqrt{209}-17}{10}\approx -3.145683229$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -4,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+4\right), the least common multiple of x+4,x.

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

To find the opposite of x+4, find the opposite of each term.

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

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

$$3x-4=5x\left(x+4\right)$$

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

$$3x-4=5x^{2}+20x$$

Subtract 5x^{2} from both sides.

$$3x-4-5x^{2}=20x$$

Subtract 20x from both sides.

$$3x-4-5x^{2}-20x=0$$

Combine 3x and -20x to get -17x.

$$-17x-4-5x^{2}=0$$

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.

$$-5x^{2}-17x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, -17 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\left(-5\right)\left(-4\right)}}{2\left(-5\right)}$$

Square -17.

$$x=\frac{-\left(-17\right)±\sqrt{289-4\left(-5\right)\left(-4\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-\left(-17\right)±\sqrt{289+20\left(-4\right)}}{2\left(-5\right)}$$

Multiply 20 times -4.

$$x=\frac{-\left(-17\right)±\sqrt{289-80}}{2\left(-5\right)}$$

Add 289 to -80.

$$x=\frac{-\left(-17\right)±\sqrt{209}}{2\left(-5\right)}$$

The opposite of -17 is 17.

$$x=\frac{17±\sqrt{209}}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{17±\sqrt{209}}{-10}$$

Now solve the equation x=\frac{17±\sqrt{209}}{-10} when ± is plus. Add 17 to \sqrt{209}.

$$x=\frac{\sqrt{209}+17}{-10}$$

Divide 17+\sqrt{209} by -10.

$$x=\frac{-\sqrt{209}-17}{10}$$

Now solve the equation x=\frac{17±\sqrt{209}}{-10} when ± is minus. Subtract \sqrt{209} from 17.

$$x=\frac{17-\sqrt{209}}{-10}$$

Divide 17-\sqrt{209} by -10.

$$x=\frac{\sqrt{209}-17}{10}$$

The equation is now solved.

$$x=\frac{-\sqrt{209}-17}{10} x=\frac{\sqrt{209}-17}{10}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -4,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+4\right), the least common multiple of x+4,x.

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

To find the opposite of x+4, find the opposite of each term.

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

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

$$3x-4=5x\left(x+4\right)$$

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

$$3x-4=5x^{2}+20x$$

Subtract 5x^{2} from both sides.

$$3x-4-5x^{2}=20x$$

Subtract 20x from both sides.

$$3x-4-5x^{2}-20x=0$$

Combine 3x and -20x to get -17x.

$$-17x-4-5x^{2}=0$$

Add 4 to both sides. Anything plus zero gives itself.

$$-17x-5x^{2}=4$$

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.

$$-5x^{2}-17x=4$$

Divide both sides by -5.

$$\frac{-5x^{2}-17x}{-5}=\frac{4}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\left(-\frac{17}{-5}\right)x=\frac{4}{-5}$$

Divide -17 by -5.

$$x^{2}+\frac{17}{5}x=\frac{4}{-5}$$

Divide 4 by -5.

$$x^{2}+\frac{17}{5}x=-\frac{4}{5}$$

Divide \frac{17}{5}, the coefficient of the x term, by 2 to get \frac{17}{10}. Then add the square of \frac{17}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{17}{5}x+\left(\frac{17}{10}\right)^{2}=-\frac{4}{5}+\left(\frac{17}{10}\right)^{2}$$

Square \frac{17}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{17}{5}x+\frac{289}{100}=-\frac{4}{5}+\frac{289}{100}$$

Add -\frac{4}{5} to \frac{289}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{17}{5}x+\frac{289}{100}=\frac{209}{100}$$

Factor x^{2}+\frac{17}{5}x+\frac{289}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{17}{10}\right)^{2}=\frac{209}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{17}{10}\right)^{2}}=\sqrt{\frac{209}{100}}$$

Simplify.

$$x+\frac{17}{10}=\frac{\sqrt{209}}{10} x+\frac{17}{10}=-\frac{\sqrt{209}}{10}$$

Subtract \frac{17}{10} from both sides of the equation.

$$x=\frac{\sqrt{209}-17}{10} x=\frac{-\sqrt{209}-17}{10}$$