Chapter 7.2, Problem 9WE

To determine

Solve the equation $3y^2=1-y$ after re-writing it in the form $a{x}^2+bx+c=0$ .

Answer to Problem 9WE

The solution set for the quadratic equation is $\left\{\frac{-1+\sqrt{13}}{6},\frac{-1-\sqrt{13}}{6}\right\}$

Explanation of Solution

Given information: The given quadratic equation is,

  $3y^2=1-y\mathrm{....}\left(\text{Given}\right)$

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 $3y^2=1-y$ can be written in the form $a{x}^2+bx+c=0$ where $\left(a\ne 0\right)$ .

  $3y^2+y-1=\mathrm{0....}\left(\text{2}\right)$

Now if $\left\{y_1,y_2\right\}$ are the roots of quadratic equation (2) then with the help of (1) $\left\{y_1,y_2\right\}$ are evaluated as,

  $\begin{array}{c}\left\{y_1,y_2\right\}=\frac{-\left(1\right)\pm \sqrt{{\left(1\right)}^2-4\left(3\right)\left(-1\right)}}{2\left(3\right)}\\ =\frac{-1\pm \sqrt{13}}{6}\mathrm{....}\left(\text{Simplified}\right)\\ =\frac{-1+\sqrt{13}}{6},\frac{-1-\sqrt{13}}{6}\end{array}$