Solve (1/x-1+1/x+2=5/4)

$$( \frac { 1 } { x - 1 } + \frac { 1 } { x + 2 } = \frac { 5 } { 4 } )$$

Solve for x

$$x = -\frac{7}{5} = -1\frac{2}{5} = -1.4$$

$$x=2$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -2,1 since division by zero is not defined. Multiply both sides of the equation by 4\left(x-1\right)\left(x+2\right), the least common multiple of x-1,x+2,4.

$$4x+8+4x-4=5\left(x-1\right)\left(x+2\right)$$

Combine 4x and 4x to get 8x.

$$8x+8-4=5\left(x-1\right)\left(x+2\right)$$

Subtract 4 from 8 to get 4.

$$8x+4=5\left(x-1\right)\left(x+2\right)$$

Use the distributive property to multiply 5 by x-1.

$$8x+4=\left(5x-5\right)\left(x+2\right)$$

Use the distributive property to multiply 5x-5 by x+2 and combine like terms.

$$8x+4=5x^{2}+5x-10$$

Subtract 5x^{2} from both sides.

$$8x+4-5x^{2}=5x-10$$

Subtract 5x from both sides.

$$8x+4-5x^{2}-5x=-10$$

Combine 8x and -5x to get 3x.

$$3x+4-5x^{2}=-10$$

Add 10 to both sides.

$$3x+4-5x^{2}+10=0$$

Add 4 and 10 to get 14.

$$3x+14-5x^{2}=0$$

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.

$$-5x^{2}+3x+14=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 3 for b, and 14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-5\right)\times 14}}{2\left(-5\right)}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-5\right)\times 14}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-3±\sqrt{9+20\times 14}}{2\left(-5\right)}$$

Multiply 20 times 14.

$$x=\frac{-3±\sqrt{9+280}}{2\left(-5\right)}$$

Add 9 to 280.

$$x=\frac{-3±\sqrt{289}}{2\left(-5\right)}$$

Take the square root of 289.

$$x=\frac{-3±17}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-3±17}{-10}$$

Now solve the equation x=\frac{-3±17}{-10} when ± is plus. Add -3 to 17.

$$x=\frac{14}{-10}$$

Reduce the fraction \frac{14}{-10} to lowest terms by extracting and canceling out 2.

$$x=-\frac{7}{5}$$

Now solve the equation x=\frac{-3±17}{-10} when ± is minus. Subtract 17 from -3.

$$x=-\frac{20}{-10}$$

Divide -20 by -10.

$$x=2$$

The equation is now solved.

$$x=-\frac{7}{5} x=2$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -2,1 since division by zero is not defined. Multiply both sides of the equation by 4\left(x-1\right)\left(x+2\right), the least common multiple of x-1,x+2,4.

$$4x+8+4x-4=5\left(x-1\right)\left(x+2\right)$$

Combine 4x and 4x to get 8x.

$$8x+8-4=5\left(x-1\right)\left(x+2\right)$$

Subtract 4 from 8 to get 4.

$$8x+4=5\left(x-1\right)\left(x+2\right)$$

Use the distributive property to multiply 5 by x-1.

$$8x+4=\left(5x-5\right)\left(x+2\right)$$

Use the distributive property to multiply 5x-5 by x+2 and combine like terms.

$$8x+4=5x^{2}+5x-10$$

Subtract 5x^{2} from both sides.

$$8x+4-5x^{2}=5x-10$$

Subtract 5x from both sides.

$$8x+4-5x^{2}-5x=-10$$

Combine 8x and -5x to get 3x.

$$3x+4-5x^{2}=-10$$

Subtract 4 from both sides.

$$3x-5x^{2}=-10-4$$

Subtract 4 from -10 to get -14.

$$3x-5x^{2}=-14$$

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.

$$-5x^{2}+3x=-14$$

Divide both sides by -5.

$$\frac{-5x^{2}+3x}{-5}=-\frac{14}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{3}{-5}x=-\frac{14}{-5}$$

Divide 3 by -5.

$$x^{2}-\frac{3}{5}x=-\frac{14}{-5}$$

Divide -14 by -5.

$$x^{2}-\frac{3}{5}x=\frac{14}{5}$$

Divide -\frac{3}{5}, the coefficient of the x term, by 2 to get -\frac{3}{10}. Then add the square of -\frac{3}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{5}x+\left(-\frac{3}{10}\right)^{2}=\frac{14}{5}+\left(-\frac{3}{10}\right)^{2}$$

Square -\frac{3}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{5}x+\frac{9}{100}=\frac{14}{5}+\frac{9}{100}$$

Add \frac{14}{5} to \frac{9}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{5}x+\frac{9}{100}=\frac{289}{100}$$

Factor x^{2}-\frac{3}{5}x+\frac{9}{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}.

$$\left(x-\frac{3}{10}\right)^{2}=\frac{289}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{10}\right)^{2}}=\sqrt{\frac{289}{100}}$$

Simplify.

$$x-\frac{3}{10}=\frac{17}{10} x-\frac{3}{10}=-\frac{17}{10}$$

Add \frac{3}{10} to both sides of the equation.

$$x=2 x=-\frac{7}{5}$$