Chapter 7.7, Problem 2WE

To determine

Find the solution for the equation $x^2-6x+8=0$ and check answer by using the theorem about the sum and product of the root.

Answer to Problem 2WE

The solution for the quadratic equation is $\left(x\right)$ equal to $\left\{4,2\right\}$ .

Explanation of Solution

Given information: The given equation is,

  $x^2-6x+8=0\mathrm{....}\left(1\right)$

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

  $\begin{array}{c}x=\frac{-\left(-6\right)\pm \sqrt{{\left(-6\right)}^2-4\left(1\right)\left(8\right)}}{2\left(1\right)}\mathrm{....}\left(\begin{array}{l}\text{Quadratic formula for}\\ a=1,b=-6,c=8\end{array}\right)\\ =\frac{6\pm \sqrt{4}}{2}\\ =\frac{6\pm 2}{2}\\ =4,2\end{array}$

Hence the solution for the quadratic equation is $\left(x\right)$ equal to $\left\{4,2\right\}$ .

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

  $\begin{array}{c}\text{Sum of roots}=\left(4\right)+\left(2\right)\\ =6\\ \text{Product of roots}=\left(4\right)\left(2\right)\\ =8\end{array}$

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

  $\begin{array}{c}\text{Sum of roots}=-\frac{\left(-6\right)}{\left(1\right)}\mathrm{....}\left(\begin{array}{l}a=1,b=-6\\ \text{and }c=8\end{array}\right)\\ =6\\ \text{Product of roots}=\frac{\left(8\right)}{\left(1\right)}\mathrm{....}\left(\begin{array}{l}a=1,b=-6\\ \text{and }c=8\end{array}\right)\\ =8\end{array}$

Hence, verified that the equation $x^2-6x+8=0$ has sum of roots 6 and product of roots 8.