Chapter 7.7, Problem 5WE

To determine

Find the solution for the equation $2x^2+2=3x$ and check answer by using the theorem about the sum and product of the root.

Answer to Problem 5WE

The solution for the quadratic equation is $\left(x\right)$ equal to $\left\{\frac{3+\sqrt{7}i}{4},\frac{3-\sqrt{7}i}{4}\right\}$ .

Explanation of Solution

Given information: The given equation can be written as,

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

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

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

Hence the solution for the quadratic equation is $\left(x\right)$ equal to $\left\{\frac{3+\sqrt{7}i}{4},\frac{3-\sqrt{7}i}{4}\right\}$ .

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

  $\begin{array}{c}\text{Sum of roots}=\left(\frac{3+\sqrt{7}i}{4}\right)+\left(\frac{3-\sqrt{7}i}{4}\right)\\ =\frac{3}{2}\\ \text{Product of roots}=\left(\frac{3+\sqrt{7}i}{4}\right)\left(\frac{3-\sqrt{7}i}{4}\right)\\ =\frac{9-7i^2}{16}\\ =\frac{9-7\left(-1\right)}{16}\\ =1\end{array}$

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

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

Hence, verified that the equation $2x^2+2=3x$ has sum of roots $\frac{3}{2}$ and product of roots $1$ .