Chapter 9.5, Problem 77E

To determine

To calculate: The sum and product of the roots of a quadratic equation. Also, find the quadratic equation whose solutions have a sum of 2 and a product of $\frac{1}{2}$ .

Answer to Problem 77E

The required sum and product of the roots are $-\frac{b}{a}$ and $\frac{c}{a}$ respectively and the required equation is $y=2x^2-4x+1$ .

Explanation of Solution

Given information:

The roots of the equation = $\frac{-b+\sqrt{b^2-4ac}}{2a}$ , $\frac{-b-\sqrt{b^2-4ac}}{2a}$

The given roots are $\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $\frac{-b-\sqrt{b^2-4ac}}{2a}$ .

Write the sum of the roots as:

  $\begin{array}{c}\frac{-b+\sqrt{b^2-4ac}}{2a}+\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{-b+\cancel{\sqrt{b^2-4ac}}-b-\cancel{\sqrt{b^2-4ac}}}{2a}\\ =\frac{-b-b}{2a}\\ =\frac{-\cancel{2}b}{\cancel{2}a}\\ =-\frac{b}{a}\end{array}$

Also, write the product of the roots as:

  $\begin{array}{c}\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)\left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right)=\left(\frac{{\left(-b\right)}^2-{\left(\sqrt{b^2-4ac}\right)}^2}{{\left(2a\right)}^2}\right)\\ =\frac{\cancel{b^2}-\cancel{b^2}+4ac}{4a^2}\\ =\frac{4ac}{4a^2}\\ =\frac{c}{a}\end{array}$

Now, the given sum and product of the roots are 2 and $\frac{1}{2}$ respectively.

So, $\begin{array}{c}-\frac{b}{a}=2\\ b=-2a\end{array}$

And,

  $\begin{array}{c}\frac{c}{a}=\frac{1}{2}\\ c=\frac{a}{2}\end{array}$

On simplifying the above expression, we can the values a , b and c are 2, $-4$ and 1 respectively.

Substitute 2 for a , $-4$ for b and 1 for c in the quadratic equation $y=a{x}^2+bx+c$ .

  $y=2x^2-4x+1$

Hence, the required sum and product of the roots are $-\frac{b}{a}$ and $\frac{c}{a}$ respectively and the required equation is $y=2x^2-4x+1$ .