Chapter 7.2, Problem 16WE

To determine

Solve the equation $\frac{t^2}{2}+1=\frac{t}{5}$ after re-writing it in the form $a{x}^2+bx+c=0$ .

Answer to Problem 16WE

The solution set for the quadratic equation is $\left\{\frac{1+7i}{5},\frac{1-7i}{5}\right\}$

Explanation of Solution

Given information: The given quadratic equation is,

  $\frac{t^2}{2}+1=\frac{t}{5}\mathrm{....}\left(\text{Given}\right)$

Formula used: If $\left(\alpha ,\beta \right)$ are the roots of the quadratic equation $a{x}^2+bx+c=0$ where $\left(a\ne 0\right)$ then,

  $\alpha ,\beta =\frac{-b\pm \sqrt{b^2-4ac}}{2a}\mathrm{....}\left(1\right)$

Calculation: The given equation $\frac{t^2}{2}+1=\frac{t}{5}$ can be written in the form $a{x}^2+bx+c=0$ where $\left(a\ne 0\right)$ .

  $\begin{array}{c}\frac{t^2}{2}+1=\frac{t}{5}\mathrm{....}\left(\text{Given}\right)\\ 5t^2+10=2t\mathrm{....}\left(\text{Multiply 10 both sides}\right)\\ \text{or, }5t^2-2t+10=\mathrm{0....}\left(2\right)\end{array}$

Now if $\left\{t_1,t_2\right\}$ are the roots of quadratic equation (2) then with the help of (1) $\left\{t_1,t_2\right\}$ are evaluated as,

  $\begin{array}{c}\left\{t_1,t_2\right\}=\frac{-\left(-2\right)\pm \sqrt{{\left(-2\right)}^2-4\left(5\right)\left(10\right)}}{2\left(5\right)}\\ =\frac{2\pm \sqrt{-196}}{10}\mathrm{....}\left(\text{Simplified}\right)\\ =\frac{2\pm 14i}{10}\mathrm{....}\left(i=\sqrt{-1}\right)\\ =\frac{1\pm 7i}{5}\\ =\frac{1+7i}{5},\frac{1-7i}{5}\end{array}$