Solve 4/m+3+m/m-4=-28/m^2-1m-12

$$\frac { 4 } { m + 3 } + \frac { m } { m - 4 } = \frac { - 28 } { m ^ { 2 } - 1 m - 12 }$$

Solve for m

$$m=-4$$

Steps Using Factoring

Variable m cannot be equal to any of the values -3,4 since division by zero is not defined. Multiply both sides of the equation by \left(m-4\right)\left(m+3\right), the least common multiple of m+3,m-4,m^{2}-m-12.

$$\left(m-4\right)\times 4+\left(m+3\right)m=-28$$

Use the distributive property to multiply m-4 by 4.

$$4m-16+\left(m+3\right)m=-28$$

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

$$4m-16+m^{2}+3m=-28$$

Combine 4m and 3m to get 7m.

$$7m-16+m^{2}=-28$$

Add 28 to both sides.

$$7m-16+m^{2}+28=0$$

Add -16 and 28 to get 12.

$$7m+12+m^{2}=0$$

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

$$m^{2}+7m+12=0$$

To solve the equation, factor m^{2}+7m+12 using formula m^{2}+\left(a+b\right)m+ab=\left(m+a\right)\left(m+b\right). To find a and b, set up a system to be solved.

$$a+b=7 ab=12$$

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 12.

$$1,12 2,6 3,4$$

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 7.

$$a=3 b=4$$

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

$$\left(m+3\right)\left(m+4\right)$$

To find equation solutions, solve m+3=0 and m+4=0.

$$m=-3 m=-4$$

Variable m cannot be equal to -3.

$$m=-4$$

Steps Using Factoring By Grouping

Variable m cannot be equal to any of the values -3,4 since division by zero is not defined. Multiply both sides of the equation by \left(m-4\right)\left(m+3\right), the least common multiple of m+3,m-4,m^{2}-m-12.

$$\left(m-4\right)\times 4+\left(m+3\right)m=-28$$

Use the distributive property to multiply m-4 by 4.

$$4m-16+\left(m+3\right)m=-28$$

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

$$4m-16+m^{2}+3m=-28$$

Combine 4m and 3m to get 7m.

$$7m-16+m^{2}=-28$$

Add 28 to both sides.

$$7m-16+m^{2}+28=0$$

Add -16 and 28 to get 12.

$$7m+12+m^{2}=0$$

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

$$m^{2}+7m+12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as m^{2}+am+bm+12. To find a and b, set up a system to be solved.

$$a+b=7 ab=1\times 12=12$$

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 12.

$$1,12 2,6 3,4$$

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 7.

$$a=3 b=4$$

Rewrite m^{2}+7m+12 as \left(m^{2}+3m\right)+\left(4m+12\right).

$$\left(m^{2}+3m\right)+\left(4m+12\right)$$

Factor out m in the first and 4 in the second group.

$$m\left(m+3\right)+4\left(m+3\right)$$

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

$$\left(m+3\right)\left(m+4\right)$$

To find equation solutions, solve m+3=0 and m+4=0.

$$m=-3 m=-4$$

Variable m cannot be equal to -3.

$$m=-4$$

Steps Using the Quadratic Formula

Variable m cannot be equal to any of the values -3,4 since division by zero is not defined. Multiply both sides of the equation by \left(m-4\right)\left(m+3\right), the least common multiple of m+3,m-4,m^{2}-m-12.

$$\left(m-4\right)\times 4+\left(m+3\right)m=-28$$

Use the distributive property to multiply m-4 by 4.

$$4m-16+\left(m+3\right)m=-28$$

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

$$4m-16+m^{2}+3m=-28$$

Combine 4m and 3m to get 7m.

$$7m-16+m^{2}=-28$$

Add 28 to both sides.

$$7m-16+m^{2}+28=0$$

Add -16 and 28 to get 12.

$$7m+12+m^{2}=0$$

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.

$$m^{2}+7m+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 7 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$m=\frac{-7±\sqrt{7^{2}-4\times 12}}{2}$$

Square 7.

$$m=\frac{-7±\sqrt{49-4\times 12}}{2}$$

Multiply -4 times 12.

$$m=\frac{-7±\sqrt{49-48}}{2}$$

Add 49 to -48.

$$m=\frac{-7±\sqrt{1}}{2}$$

Take the square root of 1.

$$m=\frac{-7±1}{2}$$

Now solve the equation m=\frac{-7±1}{2} when ± is plus. Add -7 to 1.

$$m=-\frac{6}{2}$$

Divide -6 by 2.

$$m=-3$$

Now solve the equation m=\frac{-7±1}{2} when ± is minus. Subtract 1 from -7.

$$m=-\frac{8}{2}$$

Divide -8 by 2.

$$m=-4$$

The equation is now solved.

$$m=-3 m=-4$$

Variable m cannot be equal to -3.

$$m=-4$$

Steps for Completing the Square

Variable m cannot be equal to any of the values -3,4 since division by zero is not defined. Multiply both sides of the equation by \left(m-4\right)\left(m+3\right), the least common multiple of m+3,m-4,m^{2}-m-12.

$$\left(m-4\right)\times 4+\left(m+3\right)m=-28$$

Use the distributive property to multiply m-4 by 4.

$$4m-16+\left(m+3\right)m=-28$$

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

$$4m-16+m^{2}+3m=-28$$

Combine 4m and 3m to get 7m.

$$7m-16+m^{2}=-28$$

Add 16 to both sides.

$$7m+m^{2}=-28+16$$

Add -28 and 16 to get -12.

$$7m+m^{2}=-12$$

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.

$$m^{2}+7m=-12$$

Divide 7, the coefficient of the x term, by 2 to get \frac{7}{2}. Then add the square of \frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$m^{2}+7m+\left(\frac{7}{2}\right)^{2}=-12+\left(\frac{7}{2}\right)^{2}$$

Square \frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$m^{2}+7m+\frac{49}{4}=-12+\frac{49}{4}$$

Add -12 to \frac{49}{4}.

$$m^{2}+7m+\frac{49}{4}=\frac{1}{4}$$

Factor m^{2}+7m+\frac{49}{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}.

$$\left(m+\frac{7}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(m+\frac{7}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$m+\frac{7}{2}=\frac{1}{2} m+\frac{7}{2}=-\frac{1}{2}$$

Subtract \frac{7}{2} from both sides of the equation.

$$m=-3 m=-4$$

Variable m cannot be equal to -3.

$$m=-4$$