Chapter 7.3, Problem 39WE

(a)

To determine

Solve the equation and verify C’s results.

Answer to Problem 39WE

The roots are $\alpha =\frac{5+i\sqrt{31}}{4};\beta =\frac{5-i\sqrt{31}}{4}$

Explanation of Solution

Given:

C solved the equation $2x^2-5x+7=0$ and noticed that the sum of the roots was $\frac{5}{2}$ and the product of the roots was $\frac{7}{2}$ .

Formula Used :

Roots of the standard quadratic equation $a{x}^2+bx+c=0\text{ are }x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

Calculation:

Given equation is $2x^2-5x+7=0$ .

Compare it with the standard quadratic equation , a = 2 , b = -5 , c = 7.

So, the roots are :

  $\begin{array}{l}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \Rightarrow x=\frac{-(-5)\pm \sqrt{{(-5)}^2-4(2)(7)}}{2(2)}\\ \Rightarrow x=\frac{5\pm \sqrt{25-56}}{4}\\ \Rightarrow x=\frac{5\pm i\sqrt{31}}{4}\mathrm{.................}[i=\sqrt{-1}]\end{array}$

So, the roots are $\alpha =\frac{5+i\sqrt{31}}{4};\beta =\frac{5-i\sqrt{31}}{4}$

Now, sum of the roots :

  $\begin{array}{l}\alpha +\beta =\frac{5+i\sqrt{31}}{4}+\frac{5-i\sqrt{31}}{4}\\ \Rightarrow \alpha +\beta =\frac{10}{4}\\ \Rightarrow \alpha +\beta =\frac{5}{2}\end{array}$

Product of the roots :

  $\begin{array}{l}\alpha \beta =\frac{5+i\sqrt{31}}{4}\times \frac{5-i\sqrt{31}}{4}\\ \Rightarrow \alpha \beta =\frac{25+31}{16}\mathrm{.....................}[(a+b)(a-b)=a^2-b^2;i^2=-1]\\ \Rightarrow \alpha \beta =\frac{56}{16}\\ \Rightarrow \alpha \beta =\frac{7}{2}\end{array}$

Hence, C’s calculations are verified.

(b)

To determine

Find the sum and product of the roots of given equation.

Answer to Problem 39WE

Sum of the roots = $\alpha +\beta =\frac{-13}{2}$

Product of the roots = $\alpha \beta =\frac{11}{2}$

Explanation of Solution

Given:

The equation is $2x^2+13x+11=0$ .

Formula Used :

Roots of the standard quadratic equation $a{x}^2+bx+c=0\text{ are }x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

Calculation:

Given equation is $2x^2+13x+11=0$ .

Compare it with the standard quadratic equation , a = 2 , b = 13 , c = 11.

So, the roots are :

  $\begin{array}{l}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \Rightarrow x=\frac{-13\pm \sqrt{{(13)}^2-4(2)(11)}}{2(2)}\\ \Rightarrow x=\frac{-13\pm \sqrt{81}}{4}\\ \Rightarrow x=\frac{-13\pm 9}{4}\\ \Rightarrow x=\frac{-13+9}{4}=\frac{-4}{4}=-1\text{ and }x=\frac{-13-9}{4}=-\frac{22}{4}=-\frac{11}{2}\end{array}$

So, the roots are $\alpha =-1;\beta =-\frac{11}{2}$

Now, sum of the roots :

  $\begin{array}{l}\alpha +\beta =-1-\frac{11}{2}\\ \Rightarrow \alpha +\beta =\frac{-13}{2}\end{array}$

Product of the roots :

  $\begin{array}{l}\alpha \beta =(-1)\times \frac{-11}{2}\\ \Rightarrow \alpha \beta =\frac{11}{2}\end{array}$

Hence,

Sum of the roots = $\alpha +\beta =\frac{-13}{2}$

Product of the roots = $\alpha \beta =\frac{11}{2}$

(c)

To determine

On the basis of part (a) and (b) , predict the sum and product of the roots of the equation $a{x}^2+bx+c=0$

Answer to Problem 39WE

Sum of roots = $-\frac{b}{a}$

Product of roots = $\frac{c}{a}$

Explanation of Solution

Calculation:

From part (a) and (b) :

For the equation $2x^2-5x+7=0$

Sum of roots = $-\frac{(-5)}{2}$

Product of roots $\frac{7}{2}$

For the equation $2x^2+13x+11=0$ ,

Sum of the roots = $\frac{-13}{2}=-\frac{13}{2}$

Product of the roots = $\frac{11}{2}$

So, we can observe that for the equation $a{x}^2+bx+c=0$ ,

Sum of roots = $-\frac{b}{a}$

Product of roots = $\frac{c}{a}$