Question

In the following exercises, solve by using the Quadratic Formula.
$\frac{1}{3}{m}^2+\frac{1}{12}m=\frac{1}{4}$

Short answer

The solutions of the given equation are $m=-1$and $m=\frac{3}{4}$.

Step-by-step solution

Step 1. Write the quadratic equation in standard form.

The given equation is:

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

The given equation can be written as:

$\begin{array}{rcl}\frac{1}{3}{m}^2+\frac{1}{12}m-\frac{1}{4}& =& 0\\ \frac{4m^2+m-3}{12}& =& 0\\ 4m^2+m-3& =& 0\end{array}$

Step 2. Substitute the values in Quadratic Formula.

Quadratic Formula for the equation $a{x}^2+bx+c=0$is:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

In the given equation, $a=4,b=1,c=-3$. Substitute these values in Quadratic Formula.

$m=\frac{-1\pm \sqrt{(1{)}^2-4(4\left)\right(-3)}}{2\left(4\right)}$

Step 3. Simplify the value.

On simplification, we get

$\begin{array}{rcl}m& =& \frac{-1\pm \sqrt{1+48}}{8}\\ m& =& \frac{-1\pm \sqrt{49}}{8}\\ m& =& \frac{-1\pm 7}{8}\end{array}$

It can be written as:

$\begin{array}{rcl}m& =& \frac{-1+7}{8}\text{and}m=\frac{-1-7}{8}\\ m& =& \frac{6}{8}\text{and}m=\frac{-8}{8}\\ m& =& \frac{3}{4}\text{and}m=-1\end{array}$

Step 4. Conclusion.

The solutions of the given equation are $m=-1$and $m=\frac{3}{4}$.