Chapter 7.7, Problem 8WE

To determine

Find the quadratic equation with integral coefficients whose roots are $-3\text{ and }1$ .

Answer to Problem 8WE

The quadratic equation is $a\left(x^2+2x-3\right)=0$ where $a\ne 0$ .

Explanation of Solution

Given information: It is given that the roots of the quadratic equation are $-3\text{ and }1$ .

Formula used: If $\left(S\right)$ is the sum of roots of a quadratic equation in variable $\left(x\right)$ and $\left(P\right)$ is the product of its roots, then quadratic equation in variable $\left(x\right)$ in general form is written as,

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

Calculation: Sine the roots of the quadratic equation are $-3\text{ and }1$ . Therefore evaluate sum and product of roots for the quadratic equation.

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

Now, with the help of formula (1) substitute the evaluated sum and product of roots in (1) and evaluated the quadratic.

  $\begin{array}{c}a\left(x^2-Sx+P\right)=\mathrm{0....}\left(\text{From }\left(1\right)\right)\\ a\left(x^2-\left(-2\right)x+\left(-3\right)\right)=0\\ a\left(x^2+2x-3\right)=0\mathrm{....}\left(a\ne 0\right)\end{array}$