Chapter 7.7, Problem 6ST

To determine

Tofind: thequadratic equation.

Answer to Problem 6ST

The quadratic equation is $2x^2-2x-1=0$ .

Explanation of Solution

Given information:

The roots are $\frac{1-\sqrt{3}}{2},\frac{1+\sqrt{3}}{2}$ .

Calculation: to find the quadratic equation, first find the sum of roots and products of roots,

$\begin{array}{l}\text{sum of roots}=\frac{1-\sqrt{3}}{2}+\frac{1+\sqrt{3}}{2}\\ \text{sum of roots}=1\text{ [add]}\\ \text{product of roots}=\left(\frac{1-\sqrt{3}}{2}\right)\left(\frac{1+\sqrt{3}}{2}\right)\\ \text{product of roots}=\frac{1-3}{4}\text{ [multiply]}\\ \text{product of roots}=-\frac{1}{2}\end{array}$

Now, substitute in the formula and simplify,

$\begin{array}{l}{x}^2-\left(\text{sum of roots}\right)x+\left(\text{product of roots}\right)=0\\ x^2-\left(1\right)x+\left(-\frac{1}{2}\right)=0\text{ [substitute]}\\ 2x^2-2x-1=0\text{ [simplify]}\end{array}$

Thus, the quadratic equation is $2x^2-2x-1=0$ .