Solve 48/x+4+48/x-4=5

$$\frac { 48 } { x + 4 } + \frac { 48 } { x - 4 } = 5$$

Solve for x

$$x=20$$

$$x=-\frac{4}{5}=-0.8$$

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.

$$\left(x-4\right)\times 48+\left(x+4\right)\times 48=5\left(x-4\right)\left(x+4\right)$$

Use the distributive property to multiply x-4 by 48.

$$48x-192+\left(x+4\right)\times 48=5\left(x-4\right)\left(x+4\right)$$

Use the distributive property to multiply x+4 by 48.

$$48x-192+48x+192=5\left(x-4\right)\left(x+4\right)$$

Combine 48x and 48x to get 96x.

$$96x-192+192=5\left(x-4\right)\left(x+4\right)$$

Add -192 and 192 to get 0.

$$96x=5\left(x-4\right)\left(x+4\right)$$

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

$$96x=\left(5x-20\right)\left(x+4\right)$$

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

$$96x=5x^{2}-80$$

Subtract 5x^{2} from both sides.

$$96x-5x^{2}=-80$$

Add 80 to both sides.

$$96x-5x^{2}+80=0$$

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.

$$-5x^{2}+96x+80=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 96 for b, and 80 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-96±\sqrt{96^{2}-4\left(-5\right)\times 80}}{2\left(-5\right)}$$

Square 96.

$$x=\frac{-96±\sqrt{9216-4\left(-5\right)\times 80}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-96±\sqrt{9216+20\times 80}}{2\left(-5\right)}$$

Multiply 20 times 80.

$$x=\frac{-96±\sqrt{9216+1600}}{2\left(-5\right)}$$

Add 9216 to 1600.

$$x=\frac{-96±\sqrt{10816}}{2\left(-5\right)}$$

Take the square root of 10816.

$$x=\frac{-96±104}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-96±104}{-10}$$

Now solve the equation x=\frac{-96±104}{-10} when ± is plus. Add -96 to 104.

$$x=\frac{8}{-10}$$

Reduce the fraction \frac{8}{-10} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-96±104}{-10} when ± is minus. Subtract 104 from -96.

$$x=-\frac{200}{-10}$$

Divide -200 by -10.

$$x=20$$

The equation is now solved.

$$x=-\frac{4}{5} x=20$$

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.

$$\left(x-4\right)\times 48+\left(x+4\right)\times 48=5\left(x-4\right)\left(x+4\right)$$

Use the distributive property to multiply x-4 by 48.

$$48x-192+\left(x+4\right)\times 48=5\left(x-4\right)\left(x+4\right)$$

Use the distributive property to multiply x+4 by 48.

$$48x-192+48x+192=5\left(x-4\right)\left(x+4\right)$$

Combine 48x and 48x to get 96x.

$$96x-192+192=5\left(x-4\right)\left(x+4\right)$$

Add -192 and 192 to get 0.

$$96x=5\left(x-4\right)\left(x+4\right)$$

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

$$96x=\left(5x-20\right)\left(x+4\right)$$

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

$$96x=5x^{2}-80$$

Subtract 5x^{2} from both sides.

$$96x-5x^{2}=-80$$

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.

$$-5x^{2}+96x=-80$$

Divide both sides by -5.

$$\frac{-5x^{2}+96x}{-5}=-\frac{80}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{96}{-5}x=-\frac{80}{-5}$$

Divide 96 by -5.

$$x^{2}-\frac{96}{5}x=-\frac{80}{-5}$$

Divide -80 by -5.

$$x^{2}-\frac{96}{5}x=16$$

Divide -\frac{96}{5}, the coefficient of the x term, by 2 to get -\frac{48}{5}. Then add the square of -\frac{48}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{96}{5}x+\left(-\frac{48}{5}\right)^{2}=16+\left(-\frac{48}{5}\right)^{2}$$

Square -\frac{48}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{96}{5}x+\frac{2304}{25}=16+\frac{2304}{25}$$

Add 16 to \frac{2304}{25}.

$$x^{2}-\frac{96}{5}x+\frac{2304}{25}=\frac{2704}{25}$$

Factor x^{2}-\frac{96}{5}x+\frac{2304}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{48}{5}\right)^{2}=\frac{2704}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{48}{5}\right)^{2}}=\sqrt{\frac{2704}{25}}$$

Simplify.

$$x-\frac{48}{5}=\frac{52}{5} x-\frac{48}{5}=-\frac{52}{5}$$

Add \frac{48}{5} to both sides of the equation.

$$x=20 x=-\frac{4}{5}$$