Solve 5/x+2+1/x-2=4/x-2quadx=?

$$\frac { 5 } { x + 2 } + \frac { 1 } { x - 2 } = \frac { 4 } { x - 2 } \quad x = ?$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{31}i-1}{4}\approx -0.25-1.391941091i$$

$$x=\frac{-1+\sqrt{31}i}{4}\approx -0.25+1.391941091i$$

Steps Using the Quadratic Formula

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

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

Use the distributive property to multiply x-2 by 5.

$$5x-10+x+2=\left(x+2\right)\times 4x$$

Combine 5x and x to get 6x.

$$6x-10+2=\left(x+2\right)\times 4x$$

Add -10 and 2 to get -8.

$$6x-8=\left(x+2\right)\times 4x$$

Use the distributive property to multiply x+2 by 4.

$$6x-8=\left(4x+8\right)x$$

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

$$6x-8=4x^{2}+8x$$

Subtract 4x^{2} from both sides.

$$6x-8-4x^{2}=8x$$

Subtract 8x from both sides.

$$6x-8-4x^{2}-8x=0$$

Combine 6x and -8x to get -2x.

$$-2x-8-4x^{2}=0$$

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.

$$-4x^{2}-2x-8=0$$

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

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

Square -2.

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

Multiply -4 times -4.

$$x=\frac{-\left(-2\right)±\sqrt{4+16\left(-8\right)}}{2\left(-4\right)}$$

Multiply 16 times -8.

$$x=\frac{-\left(-2\right)±\sqrt{4-128}}{2\left(-4\right)}$$

Add 4 to -128.

$$x=\frac{-\left(-2\right)±\sqrt{-124}}{2\left(-4\right)}$$

Take the square root of -124.

$$x=\frac{-\left(-2\right)±2\sqrt{31}i}{2\left(-4\right)}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{31}i}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{2±2\sqrt{31}i}{-8}$$

Now solve the equation x=\frac{2±2\sqrt{31}i}{-8} when ± is plus. Add 2 to 2i\sqrt{31}.

$$x=\frac{2+2\sqrt{31}i}{-8}$$

Divide 2+2i\sqrt{31} by -8.

$$x=\frac{-\sqrt{31}i-1}{4}$$

Now solve the equation x=\frac{2±2\sqrt{31}i}{-8} when ± is minus. Subtract 2i\sqrt{31} from 2.

$$x=\frac{-2\sqrt{31}i+2}{-8}$$

Divide 2-2i\sqrt{31} by -8.

$$x=\frac{-1+\sqrt{31}i}{4}$$

The equation is now solved.

$$x=\frac{-\sqrt{31}i-1}{4} x=\frac{-1+\sqrt{31}i}{4}$$

Steps for Completing the Square

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

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

Use the distributive property to multiply x-2 by 5.

$$5x-10+x+2=\left(x+2\right)\times 4x$$

Combine 5x and x to get 6x.

$$6x-10+2=\left(x+2\right)\times 4x$$

Add -10 and 2 to get -8.

$$6x-8=\left(x+2\right)\times 4x$$

Use the distributive property to multiply x+2 by 4.

$$6x-8=\left(4x+8\right)x$$

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

$$6x-8=4x^{2}+8x$$

Subtract 4x^{2} from both sides.

$$6x-8-4x^{2}=8x$$

Subtract 8x from both sides.

$$6x-8-4x^{2}-8x=0$$

Combine 6x and -8x to get -2x.

$$-2x-8-4x^{2}=0$$

Add 8 to both sides. Anything plus zero gives itself.

$$-2x-4x^{2}=8$$

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.

$$-4x^{2}-2x=8$$

Divide both sides by -4.

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

Dividing by -4 undoes the multiplication by -4.

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

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

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

Divide 8 by -4.

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

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

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

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=-2+\frac{1}{16}$$

Add -2 to \frac{1}{16}.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=-\frac{31}{16}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{4}\right)^{2}=-\frac{31}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{-\frac{31}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{\sqrt{31}i}{4} x+\frac{1}{4}=-\frac{\sqrt{31}i}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

$$x=\frac{-1+\sqrt{31}i}{4} x=\frac{-\sqrt{31}i-1}{4}$$