Chapter 14, Problem 6STP

To determine

To Find: a quadratic equation whose roots are given.

Answer to Problem 6STP

  $J)6x^2-5x+1=0$

Explanation of Solution

Given information:Roots of a quadratic equation are $\frac{1}{2}$ and $\frac{1}{3}$ .

Formula used:quadratic equations whose roots are $\alpha$ and $\beta$ is $x^2-(\alpha +\beta )x+\alpha \beta =0$

Calculation:

  $\begin{array}{l}giventhatrootsofquadraticequationare\frac{1}{2}and\frac{1}{3}.\\ Tabe\alpha =\frac{1}{2}and\beta =\frac{1}{3},theequationis\\ x^2-(\frac{1}{2}+\frac{1}{3})x+\frac{1}{2}\cdot \frac{1}{3}=0\\ \therefore x^2-\frac{5}{6}x+\frac{1}{6}=0\\ \therefore 6x^2-5x+1=0.\end{array}$