Solve 1/x-x/x+2=3

$$\frac { 1 } { x } - \frac { x } { x + 2 } = 3$$

Solve for x

$$x=\frac{\sqrt{57}-5}{8}\approx 0.318729304$$

$$x=\frac{-\sqrt{57}-5}{8}\approx -1.568729304$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -2,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+2\right), the least common multiple of x,x+2.

$$x+2-xx=3x\left(x+2\right)$$

Multiply x and x to get x^{2}.

$$x+2-x^{2}=3x\left(x+2\right)$$

Use the distributive property to multiply 3x by x+2.

$$x+2-x^{2}=3x^{2}+6x$$

Subtract 3x^{2} from both sides.

$$x+2-x^{2}-3x^{2}=6x$$

Subtract 6x from both sides.

$$x+2-x^{2}-3x^{2}-6x=0$$

Combine x and -6x to get -5x.

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

Combine -x^{2} and -3x^{2} to get -4x^{2}.

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

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.

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

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

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

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-4\right)\times 2}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-\left(-5\right)±\sqrt{25+16\times 2}}{2\left(-4\right)}$$

Multiply 16 times 2.

$$x=\frac{-\left(-5\right)±\sqrt{25+32}}{2\left(-4\right)}$$

Add 25 to 32.

$$x=\frac{-\left(-5\right)±\sqrt{57}}{2\left(-4\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{57}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{5±\sqrt{57}}{-8}$$

Now solve the equation x=\frac{5±\sqrt{57}}{-8} when ± is plus. Add 5 to \sqrt{57}.

$$x=\frac{\sqrt{57}+5}{-8}$$

Divide 5+\sqrt{57} by -8.

$$x=\frac{-\sqrt{57}-5}{8}$$

Now solve the equation x=\frac{5±\sqrt{57}}{-8} when ± is minus. Subtract \sqrt{57} from 5.

$$x=\frac{5-\sqrt{57}}{-8}$$

Divide 5-\sqrt{57} by -8.

$$x=\frac{\sqrt{57}-5}{8}$$

The equation is now solved.

$$x=\frac{-\sqrt{57}-5}{8} x=\frac{\sqrt{57}-5}{8}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -2,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+2\right), the least common multiple of x,x+2.

$$x+2-xx=3x\left(x+2\right)$$

Multiply x and x to get x^{2}.

$$x+2-x^{2}=3x\left(x+2\right)$$

Use the distributive property to multiply 3x by x+2.

$$x+2-x^{2}=3x^{2}+6x$$

Subtract 3x^{2} from both sides.

$$x+2-x^{2}-3x^{2}=6x$$

Subtract 6x from both sides.

$$x+2-x^{2}-3x^{2}-6x=0$$

Combine x and -6x to get -5x.

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

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

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

Combine -x^{2} and -3x^{2} to get -4x^{2}.

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

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.

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

Divide both sides by -4.

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

Dividing by -4 undoes the multiplication by -4.

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

Divide -5 by -4.

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

Reduce the fraction \frac{-2}{-4} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{5}{4}x=\frac{1}{2}$$

Divide \frac{5}{4}, the coefficient of the x term, by 2 to get \frac{5}{8}. Then add the square of \frac{5}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{4}x+\left(\frac{5}{8}\right)^{2}=\frac{1}{2}+\left(\frac{5}{8}\right)^{2}$$

Square \frac{5}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=\frac{1}{2}+\frac{25}{64}$$

Add \frac{1}{2} to \frac{25}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=\frac{57}{64}$$

Factor x^{2}+\frac{5}{4}x+\frac{25}{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}.

$$\left(x+\frac{5}{8}\right)^{2}=\frac{57}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{8}\right)^{2}}=\sqrt{\frac{57}{64}}$$

Simplify.

$$x+\frac{5}{8}=\frac{\sqrt{57}}{8} x+\frac{5}{8}=-\frac{\sqrt{57}}{8}$$

Subtract \frac{5}{8} from both sides of the equation.

$$x=\frac{\sqrt{57}-5}{8} x=\frac{-\sqrt{57}-5}{8}$$