Chapter 7.7, Problem 3WE

To determine

Find the solution for the equation $2y^2-y-6=0$ and check answer by using the theorem about the sum and product of the root.

Answer to Problem 3WE

The solution for the quadratic equation is $\left(y\right)$ equal to $\left\{2,-\frac{3}{2}\right\}$ .

Explanation of Solution

Given information: The given equation is,

  $2y^2-y-6=0\mathrm{....}\left(1\right)$

Calculation: With the help of quadratic formula the equal (1) is solved as,

  $\begin{array}{c}y=\frac{-\left(-1\right)\pm \sqrt{{\left(-1\right)}^2-4\left(2\right)\left(-6\right)}}{2\left(2\right)}\mathrm{....}\left(\begin{array}{l}\text{Quadratic formula for}\\ a=2,b=-1,c=-6\end{array}\right)\\ =\frac{1\pm \sqrt{49}}{4}\\ =\frac{1\pm 7}{4}\\ =2,-\frac{3}{2}\end{array}$

Hence the solution for the quadratic equation is $\left(y\right)$ equal to $\left\{2,-\frac{3}{2}\right\}$ .

Now evaluate, the sum $\left(S\right)$ and product $\left(P\right)$ of roots,

  $\begin{array}{c}\text{Sum of roots}=\left(2\right)+\left(-\frac{3}{2}\right)\\ =\frac{1}{2}\\ \text{Product of roots}=\left(2\right)\left(-\frac{3}{2}\right)\\ =-3\end{array}$

Again from (1) apply theorem property for sum and product of roots gives,

  $\begin{array}{c}\text{Sum of roots}=-\frac{\left(-1\right)}{\left(2\right)}\mathrm{....}\left(\begin{array}{l}a=2,b=-1\\ \text{and }c=-6\end{array}\right)\\ =\frac{1}{2}\\ \text{Product of roots}=\frac{\left(-6\right)}{\left(2\right)}\mathrm{....}\left(\begin{array}{l}a=2,b=-1\\ \text{and }c=-6\end{array}\right)\\ =-3\end{array}$

Hence, verified that the equation $2y^2-y-6=0$ has sum of roots $\frac{1}{2}$ and product of roots $-3$ .