Solve 9+3m=m^2-9/m

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

Solve for m

$$m = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$m=-3$$

Steps Using Factoring By Grouping

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

$$m\times 9+3mm=m^{2}-9$$

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

$$m\times 9+3m^{2}=m^{2}-9$$

Subtract m^{2} from both sides.

$$m\times 9+3m^{2}-m^{2}=-9$$

Combine 3m^{2} and -m^{2} to get 2m^{2}.

$$m\times 9+2m^{2}=-9$$

Add 9 to both sides.

$$m\times 9+2m^{2}+9=0$$

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

$$2m^{2}+9m+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2m^{2}+am+bm+9. To find a and b, set up a system to be solved.

$$a+b=9 ab=2\times 9=18$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 18.

$$1,18 2,9 3,6$$

Calculate the sum for each pair.

$$1+18=19 2+9=11 3+6=9$$

The solution is the pair that gives sum 9.

$$a=3 b=6$$

Rewrite 2m^{2}+9m+9 as \left(2m^{2}+3m\right)+\left(6m+9\right).

$$\left(2m^{2}+3m\right)+\left(6m+9\right)$$

Factor out m in the first and 3 in the second group.

$$m\left(2m+3\right)+3\left(2m+3\right)$$

Factor out common term 2m+3 by using distributive property.

$$\left(2m+3\right)\left(m+3\right)$$

To find equation solutions, solve 2m+3=0 and m+3=0.

$$m=-\frac{3}{2} m=-3$$

Steps Using the Quadratic Formula

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

$$m\times 9+3mm=m^{2}-9$$

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

$$m\times 9+3m^{2}=m^{2}-9$$

Subtract m^{2} from both sides.

$$m\times 9+3m^{2}-m^{2}=-9$$

Combine 3m^{2} and -m^{2} to get 2m^{2}.

$$m\times 9+2m^{2}=-9$$

Add 9 to both sides.

$$m\times 9+2m^{2}+9=0$$

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.

$$2m^{2}+9m+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 9 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$m=\frac{-9±\sqrt{9^{2}-4\times 2\times 9}}{2\times 2}$$

Square 9.

$$m=\frac{-9±\sqrt{81-4\times 2\times 9}}{2\times 2}$$

Multiply -4 times 2.

$$m=\frac{-9±\sqrt{81-8\times 9}}{2\times 2}$$

Multiply -8 times 9.

$$m=\frac{-9±\sqrt{81-72}}{2\times 2}$$

Add 81 to -72.

$$m=\frac{-9±\sqrt{9}}{2\times 2}$$

Take the square root of 9.

$$m=\frac{-9±3}{2\times 2}$$

Multiply 2 times 2.

$$m=\frac{-9±3}{4}$$

Now solve the equation m=\frac{-9±3}{4} when ± is plus. Add -9 to 3.

$$m=-\frac{6}{4}$$

Reduce the fraction \frac{-6}{4} to lowest terms by extracting and canceling out 2.

$$m=-\frac{3}{2}$$

Now solve the equation m=\frac{-9±3}{4} when ± is minus. Subtract 3 from -9.

$$m=-\frac{12}{4}$$

Divide -12 by 4.

$$m=-3$$

The equation is now solved.

$$m=-\frac{3}{2} m=-3$$

Steps for Completing the Square

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

$$m\times 9+3mm=m^{2}-9$$

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

$$m\times 9+3m^{2}=m^{2}-9$$

Subtract m^{2} from both sides.

$$m\times 9+3m^{2}-m^{2}=-9$$

Combine 3m^{2} and -m^{2} to get 2m^{2}.

$$m\times 9+2m^{2}=-9$$

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.

$$2m^{2}+9m=-9$$

Divide both sides by 2.

$$\frac{2m^{2}+9m}{2}=-\frac{9}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$m^{2}+\frac{9}{2}m=-\frac{9}{2}$$

Divide \frac{9}{2}, the coefficient of the x term, by 2 to get \frac{9}{4}. Then add the square of \frac{9}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$m^{2}+\frac{9}{2}m+\left(\frac{9}{4}\right)^{2}=-\frac{9}{2}+\left(\frac{9}{4}\right)^{2}$$

Square \frac{9}{4} by squaring both the numerator and the denominator of the fraction.

$$m^{2}+\frac{9}{2}m+\frac{81}{16}=-\frac{9}{2}+\frac{81}{16}$$

Add -\frac{9}{2} to \frac{81}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$m^{2}+\frac{9}{2}m+\frac{81}{16}=\frac{9}{16}$$

Factor m^{2}+\frac{9}{2}m+\frac{81}{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}.

$$\left(m+\frac{9}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(m+\frac{9}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$m+\frac{9}{4}=\frac{3}{4} m+\frac{9}{4}=-\frac{3}{4}$$

Subtract \frac{9}{4} from both sides of the equation.

$$m=-\frac{3}{2} m=-3$$