Chapter 1.8, Problem 15E

Solution

With the help of quadratic formula solve the equation $s^2=-14-3s$ for $\left(s\right)$ .

The solution for the quadratic equation is $s=\left\{\frac{-3+i\sqrt{47}}{2},\frac{-3-i\sqrt{47}}{2}\right\}$ .

Given:

The quadratic equation is $s^2=-14-3s$ .

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 $s^2=-14-3s$ is re-written as:

  $\begin{array}{c}{s}^2=-14-3s\mathrm{......}\left(\text{Given}\right)\\ \text{or, }{s}^2+3s+14=0\mathrm{......}\left(2\right)\end{array}$

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

  $\begin{array}{c}s=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\mathrm{......}\left(\text{From }\left(1\right)\right)\\ =\frac{-\left(3\right)\pm \sqrt{{\left(3\right)}^2-4\left(1\right)\left(14\right)}}{2\left(1\right)}\mathrm{......}\left(\because s^2+3s+14=0\right)\\ =\frac{-3\pm i\sqrt{47}}{2}\\ =\left\{\frac{-3+i\sqrt{47}}{2},\frac{-3-i\sqrt{47}}{2}\right\}\end{array}$

Hence the solution for the quadratic equation is $s=\left\{\frac{-3+i\sqrt{47}}{2},\frac{-3-i\sqrt{47}}{2}\right\}$ .