Solve 8/x-3/x+4=5

$$\frac{ 8 }{ x } - \frac{ 3 }{ x+4 } =5$$

Solve for x

$$x=\frac{\sqrt{865}}{10}-\frac{3}{2}\approx 1.441088234$$

$$x=-\frac{\sqrt{865}}{10}-\frac{3}{2}\approx -4.441088234$$

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,x+4.

$$\left(x+4\right)\times 8-x\times 3=5x\left(x+4\right)$$

Use the distributive property to multiply x+4 by 8.

$$8x+32-x\times 3=5x\left(x+4\right)$$

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

$$8x+32-x\times 3=5x^{2}+20x$$

Subtract 5x^{2} from both sides.

$$8x+32-x\times 3-5x^{2}=20x$$

Subtract 20x from both sides.

$$8x+32-x\times 3-5x^{2}-20x=0$$

Combine 8x and -20x to get -12x.

$$-12x+32-x\times 3-5x^{2}=0$$

Multiply -1 and 3 to get -3.

$$-12x+32-3x-5x^{2}=0$$

Combine -12x and -3x to get -15x.

$$-15x+32-5x^{2}=0$$

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.

$$-5x^{2}-15x+32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, -15 for b, and 32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\left(-5\right)\times 32}}{2\left(-5\right)}$$

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\left(-5\right)\times 32}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-\left(-15\right)±\sqrt{225+20\times 32}}{2\left(-5\right)}$$

Multiply 20 times 32.

$$x=\frac{-\left(-15\right)±\sqrt{225+640}}{2\left(-5\right)}$$

Add 225 to 640.

$$x=\frac{-\left(-15\right)±\sqrt{865}}{2\left(-5\right)}$$

The opposite of -15 is 15.

$$x=\frac{15±\sqrt{865}}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{15±\sqrt{865}}{-10}$$

Now solve the equation x=\frac{15±\sqrt{865}}{-10} when ± is plus. Add 15 to \sqrt{865}.

$$x=\frac{\sqrt{865}+15}{-10}$$

Divide 15+\sqrt{865} by -10.

$$x=-\frac{\sqrt{865}}{10}-\frac{3}{2}$$

Now solve the equation x=\frac{15±\sqrt{865}}{-10} when ± is minus. Subtract \sqrt{865} from 15.

$$x=\frac{15-\sqrt{865}}{-10}$$

Divide 15-\sqrt{865} by -10.

$$x=\frac{\sqrt{865}}{10}-\frac{3}{2}$$

The equation is now solved.

$$x=-\frac{\sqrt{865}}{10}-\frac{3}{2} x=\frac{\sqrt{865}}{10}-\frac{3}{2}$$

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,x+4.

$$\left(x+4\right)\times 8-x\times 3=5x\left(x+4\right)$$

Use the distributive property to multiply x+4 by 8.

$$8x+32-x\times 3=5x\left(x+4\right)$$

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

$$8x+32-x\times 3=5x^{2}+20x$$

Subtract 5x^{2} from both sides.

$$8x+32-x\times 3-5x^{2}=20x$$

Subtract 20x from both sides.

$$8x+32-x\times 3-5x^{2}-20x=0$$

Combine 8x and -20x to get -12x.

$$-12x+32-x\times 3-5x^{2}=0$$

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

$$-12x-x\times 3-5x^{2}=-32$$

Multiply -1 and 3 to get -3.

$$-12x-3x-5x^{2}=-32$$

Combine -12x and -3x to get -15x.

$$-15x-5x^{2}=-32$$

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.

$$-5x^{2}-15x=-32$$

Divide both sides by -5.

$$\frac{-5x^{2}-15x}{-5}=-\frac{32}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\left(-\frac{15}{-5}\right)x=-\frac{32}{-5}$$

Divide -15 by -5.

$$x^{2}+3x=-\frac{32}{-5}$$

Divide -32 by -5.

$$x^{2}+3x=\frac{32}{5}$$

Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=\frac{32}{5}+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=\frac{32}{5}+\frac{9}{4}$$

Add \frac{32}{5} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+3x+\frac{9}{4}=\frac{173}{20}$$

Factor x^{2}+3x+\frac{9}{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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{173}{20}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{173}{20}}$$

Simplify.

$$x+\frac{3}{2}=\frac{\sqrt{865}}{10} x+\frac{3}{2}=-\frac{\sqrt{865}}{10}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=\frac{\sqrt{865}}{10}-\frac{3}{2} x=-\frac{\sqrt{865}}{10}-\frac{3}{2}$$