Chapter 1.8, Problem 20E

Solution

With the help of quadratic formula solve the equation $6-2t^2=9t+15$ for $\left(t\right)$ .

The solution for the quadratic equation is $t=\left\{-\frac{3}{2},-3\right\}$ .

Given:

The quadratic equation is $6-2t^2=9t+15$ .

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 $6-2t^2=9t+15$ is re-written as:

  $\begin{array}{c}6-2t^2=9t+15\mathrm{......}\left(\text{Given}\right)\\ \text{or, }9t+15-6+2t^2=0\\ 2t^2+9t+9=0\mathrm{......}\left(2\right)\end{array}$

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

  $\begin{array}{c}t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\mathrm{......}\left(\text{From }\left(1\right)\right)\\ =\frac{-\left(9\right)\pm \sqrt{{\left(9\right)}^2-4\left(2\right)\left(9\right)}}{2\left(2\right)}\mathrm{......}\left(\because 2t^2+9t+9=0\right)\\ =\frac{-9\pm \sqrt{9}}{4}\\ =\frac{-9\pm 3}{4}\\ =\frac{-9+3}{4},\frac{-9-3}{4}\\ =\left\{-\frac{3}{2},-3\right\}\end{array}$

Hence the solution for the quadratic equation is $t=\left\{-\frac{3}{2},-3\right\}$ .