Question

If $\alpha$ and $\beta$ are roots of the equation $a{x}^2+bx+c=0$ then the equation whose roots are $\alpha +\frac{1}{\beta }$ are $\beta +\frac{1}{\alpha }$ is

• A. $ac{x}^2+(a+c)bx+{(a+c)}^2=0$
• B. $ab{x}^2+(a+c)bx+{(a+c)}^2=0$
• C. $ac{x}^2+(a+b)cx+{(a+c)}^2=0$
• D. $ac{x}^2-(a+c)bx+{(a+c)}^2=0$

Answer 1

The correct option is A $ac{x}^2+(a+c)bx+{(a+c)}^2=0$

We have $\alpha +\beta =\frac{-b}{a}$ and $\alpha \beta =\frac{c}{a}$

Sum of the roots$=\alpha +\frac{1}{\beta }+\beta +\frac{1}{\alpha }$

$=\alpha +\beta +\frac{1}{\alpha }+\frac{1}{\beta }$

$=\alpha +\beta +\frac{\alpha +\beta }{\alpha \beta }$

$=\frac{-b}{a}+\frac{\frac{-b}{a}}{\frac{c}{a}}$

$=\frac{-b}{a}+\frac{-b}{c}$

$=\frac{-b(a+c)}{ac}$

Product of the roots$=(\alpha +\frac{1}{\beta })(\beta +\frac{1}{\alpha })$

$=\alpha \beta +\frac{\alpha }{\alpha }+\frac{\beta }{\beta }+\frac{1}{\alpha \beta }$

$=\alpha \beta +\frac{1}{\alpha \beta }+2$

$=\frac{{\left(\alpha \beta \right)}^2+1}{\alpha \beta }+2$

$=\frac{\frac{c^2}{a^2}+1}{\frac{c}{a}}+2$

$=\frac{c^2+a^2}{ac}+2$

$=\frac{c^2+a^2+2ac}{ac}$

$=\frac{{(c+a)}^2}{ac}$

Required equation $=x^2-\left(Sumoftheroots\right)x+\left(productoftheroots\right)=0$

$x^2-\left(\frac{-b(a+c)}{ac}\right)x+\left(\frac{{(c+a)}^2}{ac}\right)=0$

or $ac{x}^2+b(a+c)x+{(c+a)}^2=0$