Solve -2/x(x+2)=x+3/x+2

$$- \frac { 2 } { x ( x + 2 ) } = \frac { x + 3 } { x + 2 }$$

Solve for x

$$x=-1$$

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\left(x+2\right),x+2.

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

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

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

Swap sides so that all variable terms are on the left hand side.

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

Add 2 to both sides.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 3 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\times 2}}{2}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 2}}{2}$$

Multiply -4 times 2.

$$x=\frac{-3±\sqrt{9-8}}{2}$$

Add 9 to -8.

$$x=\frac{-3±\sqrt{1}}{2}$$

Take the square root of 1.

$$x=\frac{-3±1}{2}$$

Now solve the equation x=\frac{-3±1}{2} when ± is plus. Add -3 to 1.

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

Divide -2 by 2.

$$x=-1$$

Now solve the equation x=\frac{-3±1}{2} when ± is minus. Subtract 1 from -3.

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

Divide -4 by 2.

$$x=-2$$

The equation is now solved.

$$x=-1 x=-2$$

Variable x cannot be equal to -2.

$$x=-1$$

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\left(x+2\right),x+2.

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

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

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

Swap sides so that all variable terms are on the left hand side.

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

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

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

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=-2+\frac{9}{4}$$

Add -2 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{1}{4}$$

Factor x^{2}+3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

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

Subtract \frac{3}{2} from both sides of the equation.

$$x=-1 x=-2$$

Variable x cannot be equal to -2.

$$x=-1$$