Solve 1/x-2+1/x+4=8/7

$$\frac { 1 } { x - 2 } + \frac { 1 } { x + 4 } = \frac { 8 } { 7 }$$

Solve for x

$$x = -\frac{13}{4} = -3\frac{1}{4} = -3.25$$

$$x=3$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -4,2 since division by zero is not defined. Multiply both sides of the equation by 7\left(x-2\right)\left(x+4\right), the least common multiple of x-2,x+4,7.

$$7x+28+7x-14=8\left(x-2\right)\left(x+4\right)$$

Combine 7x and 7x to get 14x.

$$14x+28-14=8\left(x-2\right)\left(x+4\right)$$

Subtract 14 from 28 to get 14.

$$14x+14=8\left(x-2\right)\left(x+4\right)$$

Use the distributive property to multiply 8 by x-2.

$$14x+14=\left(8x-16\right)\left(x+4\right)$$

Use the distributive property to multiply 8x-16 by x+4 and combine like terms.

$$14x+14=8x^{2}+16x-64$$

Subtract 8x^{2} from both sides.

$$14x+14-8x^{2}=16x-64$$

Subtract 16x from both sides.

$$14x+14-8x^{2}-16x=-64$$

Combine 14x and -16x to get -2x.

$$-2x+14-8x^{2}=-64$$

Add 64 to both sides.

$$-2x+14-8x^{2}+64=0$$

Add 14 and 64 to get 78.

$$-2x+78-8x^{2}=0$$

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.

$$-8x^{2}-2x+78=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -8 for a, -2 for b, and 78 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-8\right)\times 78}}{2\left(-8\right)}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-8\right)\times 78}}{2\left(-8\right)}$$

Multiply -4 times -8.

$$x=\frac{-\left(-2\right)±\sqrt{4+32\times 78}}{2\left(-8\right)}$$

Multiply 32 times 78.

$$x=\frac{-\left(-2\right)±\sqrt{4+2496}}{2\left(-8\right)}$$

Add 4 to 2496.

$$x=\frac{-\left(-2\right)±\sqrt{2500}}{2\left(-8\right)}$$

Take the square root of 2500.

$$x=\frac{-\left(-2\right)±50}{2\left(-8\right)}$$

The opposite of -2 is 2.

$$x=\frac{2±50}{2\left(-8\right)}$$

Multiply 2 times -8.

$$x=\frac{2±50}{-16}$$

Now solve the equation x=\frac{2±50}{-16} when ± is plus. Add 2 to 50.

$$x=\frac{52}{-16}$$

Reduce the fraction \frac{52}{-16} to lowest terms by extracting and canceling out 4.

$$x=-\frac{13}{4}$$

Now solve the equation x=\frac{2±50}{-16} when ± is minus. Subtract 50 from 2.

$$x=-\frac{48}{-16}$$

Divide -48 by -16.

$$x=3$$

The equation is now solved.

$$x=-\frac{13}{4} x=3$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -4,2 since division by zero is not defined. Multiply both sides of the equation by 7\left(x-2\right)\left(x+4\right), the least common multiple of x-2,x+4,7.

$$7x+28+7x-14=8\left(x-2\right)\left(x+4\right)$$

Combine 7x and 7x to get 14x.

$$14x+28-14=8\left(x-2\right)\left(x+4\right)$$

Subtract 14 from 28 to get 14.

$$14x+14=8\left(x-2\right)\left(x+4\right)$$

Use the distributive property to multiply 8 by x-2.

$$14x+14=\left(8x-16\right)\left(x+4\right)$$

Use the distributive property to multiply 8x-16 by x+4 and combine like terms.

$$14x+14=8x^{2}+16x-64$$

Subtract 8x^{2} from both sides.

$$14x+14-8x^{2}=16x-64$$

Subtract 16x from both sides.

$$14x+14-8x^{2}-16x=-64$$

Combine 14x and -16x to get -2x.

$$-2x+14-8x^{2}=-64$$

Subtract 14 from both sides.

$$-2x-8x^{2}=-64-14$$

Subtract 14 from -64 to get -78.

$$-2x-8x^{2}=-78$$

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.

$$-8x^{2}-2x=-78$$

Divide both sides by -8.

$$\frac{-8x^{2}-2x}{-8}=-\frac{78}{-8}$$

Dividing by -8 undoes the multiplication by -8.

$$x^{2}+\left(-\frac{2}{-8}\right)x=-\frac{78}{-8}$$

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

$$x^{2}+\frac{1}{4}x=-\frac{78}{-8}$$

Reduce the fraction \frac{-78}{-8} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{1}{4}x=\frac{39}{4}$$

Divide \frac{1}{4}, the coefficient of the x term, by 2 to get \frac{1}{8}. Then add the square of \frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{4}x+\left(\frac{1}{8}\right)^{2}=\frac{39}{4}+\left(\frac{1}{8}\right)^{2}$$

Square \frac{1}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=\frac{39}{4}+\frac{1}{64}$$

Add \frac{39}{4} to \frac{1}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=\frac{625}{64}$$

Factor x^{2}+\frac{1}{4}x+\frac{1}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{8}\right)^{2}=\frac{625}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{8}\right)^{2}}=\sqrt{\frac{625}{64}}$$

Simplify.

$$x+\frac{1}{8}=\frac{25}{8} x+\frac{1}{8}=-\frac{25}{8}$$

Subtract \frac{1}{8} from both sides of the equation.

$$x=3 x=-\frac{13}{4}$$