Chapter 1.8, Problem 29E

Solution

Solve the equation $5p^2+40p+100=25$ by quadratic formula and then by factoring to check solution.

The solution for the quadratic equation is $p=\left\{-3,-5\right\}$ .

Given:

The quadratic equation is $5p^2+40p+100=25$ .

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 $5p^2+40p+100=25$ is re-written as:

  $\begin{array}{c}5p^2+40p+100=25\mathrm{......}\left(\text{Given}\right)\\ 5p^2+40p+75=0\\ p^2+8p+15=0\mathrm{......}\left(2\right)\end{array}$

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

  $\begin{array}{c}p=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\mathrm{......}\left(\text{From }\left(1\right)\right)\\ =\frac{-\left(8\right)\pm \sqrt{{\left(8\right)}^2-4\left(1\right)\left(15\right)}}{2\left(1\right)}\mathrm{......}\left(\because p^2+8p+15=0\right)\\ =\frac{-8\pm \sqrt{4}}{2}\\ =\frac{-8+2}{2},\frac{-8-2}{2}\\ =\left\{-3,-5\right\}\end{array}$

Hence the solution for the quadratic equation is $p=\left\{-3,-5\right\}$ .

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

  $\begin{array}{c}{p}^2+8p+15=0\mathrm{......}\left(\text{From }\left(2\right)\right)\\ p^2+3p+5p+15=0\\ p\left(p+3\right)+5\left(p+3\right)=0\\ \left(p+3\right)\left(p+5\right)=0\\ p+3=0\text{ or }p+5=0\\ p=-3\text{ or }p=-5\\ p=\left\{-3,-5\right\}\end{array}$

From both the techniques the solutions are same for of the quadratic equation. Hence the required solution is $p=\left\{-3,-5\right\}$ of quadratic equation.