Solve 1/2m^2+6m+1/2=5/2m^2+6m

$$\frac { 1 } { 2 m ^ { 2 } + 6 m } + \frac { 1 } { 2 } = \frac { 5 } { 2 m ^ { 2 } + 6 m }$$

Solve for m

$$m=-4$$

$$m=1$$

Steps Using Factoring

Variable m cannot be equal to any of the values -3,0 since division by zero is not defined. Multiply both sides of the equation by 2m\left(m+3\right), the least common multiple of 2m^{2}+6m,2.

$$1+2m\left(m+3\right)\times \frac{1}{2}=5$$

Multiply 2 and \frac{1}{2} to get 1.

$$1+m\left(m+3\right)=5$$

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

$$1+m^{2}+3m=5$$

Subtract 5 from both sides.

$$1+m^{2}+3m-5=0$$

Subtract 5 from 1 to get -4.

$$-4+m^{2}+3m=0$$

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

$$m^{2}+3m-4=0$$

To solve the equation, factor m^{2}+3m-4 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.

$$a+b=3 ab=-4$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -4.

$$-1,4 -2,2$$

Calculate the sum for each pair.

$$-1+4=3 -2+2=0$$

The solution is the pair that gives sum 3.

$$a=-1 b=4$$

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

$$\left(m-1\right)\left(m+4\right)$$

To find equation solutions, solve m-1=0 and m+4=0.

$$m=1 m=-4$$

Steps Using Factoring By Grouping

Variable m cannot be equal to any of the values -3,0 since division by zero is not defined. Multiply both sides of the equation by 2m\left(m+3\right), the least common multiple of 2m^{2}+6m,2.

$$1+2m\left(m+3\right)\times \frac{1}{2}=5$$

Multiply 2 and \frac{1}{2} to get 1.

$$1+m\left(m+3\right)=5$$

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

$$1+m^{2}+3m=5$$

Subtract 5 from both sides.

$$1+m^{2}+3m-5=0$$

Subtract 5 from 1 to get -4.

$$-4+m^{2}+3m=0$$

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

$$m^{2}+3m-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as m^{2}+am+bm-4. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\left(-4\right)=-4$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -4.

$$-1,4 -2,2$$

Calculate the sum for each pair.

$$-1+4=3 -2+2=0$$

The solution is the pair that gives sum 3.

$$a=-1 b=4$$

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

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

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

$$m\left(m-1\right)+4\left(m-1\right)$$

Factor out common term m-1 by using distributive property.

$$\left(m-1\right)\left(m+4\right)$$

To find equation solutions, solve m-1=0 and m+4=0.

$$m=1 m=-4$$

Steps Using the Quadratic Formula

Variable m cannot be equal to any of the values -3,0 since division by zero is not defined. Multiply both sides of the equation by 2m\left(m+3\right), the least common multiple of 2m^{2}+6m,2.

$$1+2m\left(m+3\right)\times \frac{1}{2}=5$$

Multiply 2 and \frac{1}{2} to get 1.

$$1+m\left(m+3\right)=5$$

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

$$1+m^{2}+3m=5$$

Subtract 5 from both sides.

$$1+m^{2}+3m-5=0$$

Subtract 5 from 1 to get -4.

$$-4+m^{2}+3m=0$$

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.

$$m^{2}+3m-4=0$$

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

$$m=\frac{-3±\sqrt{3^{2}-4\left(-4\right)}}{2}$$

Square 3.

$$m=\frac{-3±\sqrt{9-4\left(-4\right)}}{2}$$

Multiply -4 times -4.

$$m=\frac{-3±\sqrt{9+16}}{2}$$

Add 9 to 16.

$$m=\frac{-3±\sqrt{25}}{2}$$

Take the square root of 25.

$$m=\frac{-3±5}{2}$$

Now solve the equation m=\frac{-3±5}{2} when ± is plus. Add -3 to 5.

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

Divide 2 by 2.

$$m=1$$

Now solve the equation m=\frac{-3±5}{2} when ± is minus. Subtract 5 from -3.

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

Divide -8 by 2.

$$m=-4$$

The equation is now solved.

$$m=1 m=-4$$

Steps for Completing the Square

Variable m cannot be equal to any of the values -3,0 since division by zero is not defined. Multiply both sides of the equation by 2m\left(m+3\right), the least common multiple of 2m^{2}+6m,2.

$$1+2m\left(m+3\right)\times \frac{1}{2}=5$$

Multiply 2 and \frac{1}{2} to get 1.

$$1+m\left(m+3\right)=5$$

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

$$1+m^{2}+3m=5$$

Subtract 1 from both sides.

$$m^{2}+3m=5-1$$

Subtract 1 from 5 to get 4.

$$m^{2}+3m=4$$

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.

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

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

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

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

$$m^{2}+3m+\frac{9}{4}=\frac{25}{4}$$

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

$$\left(m+\frac{3}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(m+\frac{3}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$m+\frac{3}{2}=\frac{5}{2} m+\frac{3}{2}=-\frac{5}{2}$$

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

$$m=1 m=-4$$