Solve q+2/q=-3

$$q + \frac { 2 } { q } = - 3$$

Solve for q

$$q=-2$$

$$q=-1$$

Steps Using Factoring

Variable q cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by q.

$$qq+2=-3q$$

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

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

Add 3q to both sides.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor q^{2}+3q+2 using formula q^{2}+\left(a+b\right)q+ab=\left(q+a\right)\left(q+b\right). To find a and b, set up a system to be solved.

$$a+b=3 ab=2$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=2$$

Rewrite factored expression \left(q+a\right)\left(q+b\right) using the obtained values.

$$\left(q+1\right)\left(q+2\right)$$

To find equation solutions, solve q+1=0 and q+2=0.

$$q=-1 q=-2$$

Steps Using Factoring By Grouping

Variable q cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by q.

$$qq+2=-3q$$

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

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

Add 3q to both sides.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as q^{2}+aq+bq+2. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\times 2=2$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=2$$

Rewrite q^{2}+3q+2 as \left(q^{2}+q\right)+\left(2q+2\right).

$$\left(q^{2}+q\right)+\left(2q+2\right)$$

Factor out q in the first and 2 in the second group.

$$q\left(q+1\right)+2\left(q+1\right)$$

Factor out common term q+1 by using distributive property.

$$\left(q+1\right)\left(q+2\right)$$

To find equation solutions, solve q+1=0 and q+2=0.

$$q=-1 q=-2$$

Steps Using the Quadratic Formula

Variable q cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by q.

$$qq+2=-3q$$

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

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

Add 3q to both sides.

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

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.

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

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}.

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

Square 3.

$$q=\frac{-3±\sqrt{9-4\times 2}}{2}$$

Multiply -4 times 2.

$$q=\frac{-3±\sqrt{9-8}}{2}$$

Add 9 to -8.

$$q=\frac{-3±\sqrt{1}}{2}$$

Take the square root of 1.

$$q=\frac{-3±1}{2}$$

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

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

Divide -2 by 2.

$$q=-1$$

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

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

Divide -4 by 2.

$$q=-2$$

The equation is now solved.

$$q=-1 q=-2$$

Steps for Completing the Square

Variable q cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by q.

$$qq+2=-3q$$

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

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

Add 3q to both sides.

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

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

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

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.

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

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

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

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

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

Factor q^{2}+3q+\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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

$$q=-1 q=-2$$