Chapter 1.8, Problem 22E

Solution

Solve the equation $z^2+15z+24=-32$ by quadratic formula and then by factoring to check solution.

The solution for the quadratic equation is $z=\left\{-7,-8\right\}$ .

Given:

The quadratic equation is $z^2+15z+24=-32$ .

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 $z^2+15z+24=-32$ is re-written as:

  $\begin{array}{c}{z}^2+15z+24=-32\mathrm{......}\left(\text{Given}\right)\\ z^2+15z+56=0\mathrm{......}\left(2\right)\end{array}$

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

  $\begin{array}{c}z=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\mathrm{......}\left(\text{From }\left(1\right)\right)\\ =\frac{-\left(15\right)\pm \sqrt{{\left(15\right)}^2-4\left(1\right)\left(56\right)}}{2\left(1\right)}\mathrm{......}\left(\because z^2+15z+56=0\right)\\ =\frac{-15\pm \sqrt{1}}{2}\\ =\frac{-15+1}{2},\frac{-15-1}{2}\\ z=\left\{-7,-8\right\}\end{array}$

Hence the solution for the quadratic equation is $z=\left\{8,7\right\}$ .

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

  $\begin{array}{c}{z}^2+15z+56=0\mathrm{......}\left(\text{From }\left(2\right)\right)\\ z^2+8z+7z+56=0\\ z\left(z+8\right)+7\left(z+8\right)=0\\ \left(z+8\right)\left(z+7\right)=0\\ z+8=0\text{ or }z+7=0\\ z=-8\text{ or }z=-7\end{array}$

From both the techniques the solutions are same for $\left(z\right)$ . Hence the required solution is $z=\left\{-7,-8\right\}$ of quadratic equation.