Chapter 7.3, Problem 40WE

(a)

To determine

Find the roots of $9x^2-6x+2=0$

Answer to Problem 40WE

The roots are $\alpha =\frac{1+i}{3};\beta =\frac{1-i}{3}$

Explanation of Solution

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 $9x^2-6x+2=0$ .

Compare it with the standard quadratic equation , a = 9 , b = -6 , c = 2.

So, the roots are :

  $\begin{array}{l}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \Rightarrow x=\frac{-(-6)\pm \sqrt{{(-6)}^2-4(9)(2)}}{2(9)}\\ \Rightarrow x=\frac{6\pm \sqrt{36-72}}{18}\\ \Rightarrow x=\frac{6\pm i\sqrt{36}}{18}\mathrm{.................}[i=\sqrt{-1}]\\ \Rightarrow x=\frac{6\pm 6i}{18}\\ \Rightarrow x=\frac{1\pm i}{3}\end{array}$

So, the roots are $\alpha =\frac{1+i}{3};\beta =\frac{1-i}{3}$

(b)

To determine

Verify that sum and product of roots of $9x^2-6x+2=0$ are $\frac{6}{9}\text{ and }\frac{2}{9}$ .

Explanation of Solution

Calculation:

From part (a) , we have

The roots of the equation $9x^2-6x+2=0$ are $\alpha =\frac{1+i}{3};\beta =\frac{1-i}{3}$

Now, sum of the roots :

  $\begin{array}{l}\alpha +\beta =\frac{1+i}{3}+\frac{1-i}{3}\\ \Rightarrow \alpha +\beta =\frac{2}{3}=\frac{6}{9}\mathrm{.........}[\text{multiply and divide by 3]}\end{array}$

Product of the roots :

  $\begin{array}{l}\alpha \beta =\frac{1+i}{3}\times \frac{1-i}{3}\\ \Rightarrow \alpha \beta =\frac{1+1}{9}\mathrm{.....................}[(a+b)(a-b)=a^2-b^2;i^2=-1]\\ \Rightarrow \alpha \beta =\frac{2}{9}\end{array}$

Hence verified .

(c)

To determine

Find the sum and product of roots of $3x^2+7x+8=0$

Answer to Problem 40WE

Sum of roots = $-\frac{7}{3}$

Product of roots = $\frac{8}{3}$

Explanation of Solution

Calculation:

From part (a) and (b) :

For the equation $9x^2-6x+2=0$

Sum of roots = $-\frac{(-6)}{9}$

Product of roots $\frac{2}{9}$

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

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

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

Hence , for the equation $3x^2+7x+8=0$

a = 3, b = 7 , c = 8

Sum of roots = $=-\frac{b}{a}=-\frac{7}{3}$

Product of roots = $\frac{c}{a}=\frac{8}{3}$