Chapter 1.8, Problem 24E

Solution

Solve the equation $m^2+5m-99=3m$ by quadratic formula and then by factoring to check solution.

The solution for the quadratic equation is $m=\left\{-11,9\right\}$ .

Given:

The quadratic equation is $m^2+5m-99=3m$ .

Concept used:

Any quadratic equation of the form $a{x}^2+bx+c=0\text{ where }a\ne 0$ has the roots $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\mathrm{......}\left(1\right)$

Calculation:

The quadratic equation $m^2+5m-99=3m$ is re-written as:

  $\begin{array}{c}{m}^2+5m-99=3m\mathrm{......}\left(\text{Given}\right)\\ m^2+2m-99=0\mathrm{......}\left(2\right)\end{array}$

The solution for the quadratic equation $\left(2\right)$ can be evaluated with the help of formula $\left(1\right)$ . Therefore,

  $\begin{array}{c}m=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\mathrm{......}\left(\text{From }\left(1\right)\right)\\ =\frac{-\left(2\right)\pm \sqrt{{\left(2\right)}^2-4\left(1\right)\left(-99\right)}}{2\left(1\right)}\mathrm{......}\left(\because m^2+2m-99=0\right)\\ =\frac{-2\pm \sqrt{400}}{2}\\ =\frac{-2+20}{2},\frac{-2-20}{2}\\ x=\left\{9,-11\right\}\end{array}$

Hence the solution for the quadratic equation is $m=\left\{9,-11\right\}$ .

Now, factoring the equation $\left(2\right)$ gives,

  $\begin{array}{c}{m}^2+2m-99=0\mathrm{......}\left(\text{From }\left(2\right)\right)\\ m^2+11m-9m-99=0\\ m\left(m+11\right)-9\left(m+11\right)=0\\ \left(m+11\right)\left(m-9\right)=0\\ m+11=0\text{ or }m-9=0\\ m=-11\text{ or }m=9\\ m=\left\{-11,9\right\}\end{array}$

From both the techniques the solutions are same for of the quadratic equation. Hence the required solution is $m=\left\{-11,9\right\}$ of quadratic equation.