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

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

Solve for x

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

Steps Using Factoring

Variable x cannot be equal to any of the values 0,2 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-2,x,x^{2}-2x.

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

Use the distributive property to multiply x by 2x+1.

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

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

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

Combine x and 4x to get 5x.

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

Add 8 to both sides.

$$2x^{2}+5x-8+8=0$$

Add -8 and 8 to get 0.

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

Factor out x.

$$x\left(2x+5\right)=0$$

To find equation solutions, solve x=0 and 2x+5=0.

$$x=0 x=-\frac{5}{2}$$

Variable x cannot be equal to 0.

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

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,2 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-2,x,x^{2}-2x.

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

Use the distributive property to multiply x by 2x+1.

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

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

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

Combine x and 4x to get 5x.

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

Add 8 to both sides.

$$2x^{2}+5x-8+8=0$$

Add -8 and 8 to get 0.

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

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

$$x=\frac{-5±\sqrt{5^{2}}}{2\times 2}$$

Take the square root of 5^{2}.

$$x=\frac{-5±5}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-5±5}{4}$$

Now solve the equation x=\frac{-5±5}{4} when ± is plus. Add -5 to 5.

$$x=\frac{0}{4}$$

Divide 0 by 4.

$$x=0$$

Now solve the equation x=\frac{-5±5}{4} when ± is minus. Subtract 5 from -5.

$$x=-\frac{10}{4}$$

Reduce the fraction \frac{-10}{4} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

$$x=0 x=-\frac{5}{2}$$

Variable x cannot be equal to 0.

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

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,2 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-2,x,x^{2}-2x.

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

Use the distributive property to multiply x by 2x+1.

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

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

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

Combine x and 4x to get 5x.

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

Add 8 to both sides.

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

Add -8 and 8 to get 0.

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

Divide both sides by 2.

$$\frac{2x^{2}+5x}{2}=\frac{0}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{5}{2}x=\frac{0}{2}$$

Divide 0 by 2.

$$x^{2}+\frac{5}{2}x=0$$

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

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

Square \frac{5}{4} by squaring both the numerator and the denominator of the fraction.

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

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

$$\left(x+\frac{5}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract \frac{5}{4} from both sides of the equation.

$$x=0 x=-\frac{5}{2}$$

Variable x cannot be equal to 0.

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