Chapter 1.8, Problem 25E

Solution

Solve the equation $s^2-s-3=s$ by quadratic formula and then by factoring to check solution.

The solution for the quadratic equation is $s=\left\{-1,3\right\}$ .

Given:

The quadratic equation is $s^2-s-3=s$ .

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 $s^2-s-3=s$ is re-written as:

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

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

  $\begin{array}{c}s=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\mathrm{......}\left(\text{From }\left(1\right)\right)\\ =\frac{-\left(-2\right)\pm \sqrt{{\left(-2\right)}^2-4\left(1\right)\left(-3\right)}}{2\left(1\right)}\mathrm{......}\left(\because s^2-2s-3=0\right)\\ =\frac{2\pm \sqrt{16}}{2}\\ =\frac{2+4}{2},\frac{2-4}{2}\\ s=\left\{3,-1\right\}\end{array}$

Hence the solution for the quadratic equation is $s=\left\{3,-1\right\}$ .

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

  $\begin{array}{c}{s}^2-2s-3=0\mathrm{......}\left(\text{From }\left(2\right)\right)\\ s^2-3s+s-3=0\\ s\left(s-3\right)+1\left(s-3\right)=0\\ \left(s+1\right)\left(s-3\right)=0\\ s+1=0\text{ or }s-3=0\\ s=-1\text{ or }s=3\\ s=\left\{-1,3\right\}\end{array}$

From both the techniques the solutions are same for of the quadratic equation. Hence the required solution is $s=\left\{-1,3\right\}$ of quadratic equation.