Question

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

Let $\alpha$ and $\beta$ be the roots of the equation $a{x}^2+bx+c=0$ then $\alpha +\beta =\frac{-b}{a}$ and $\alpha \beta =\frac{c}{a}$

Given:$\alpha +\frac{1}{\beta }$ and $\beta +\frac{1}{\alpha }$ are the roots then

$x^2-(\alpha +\frac{1}{\beta }+\beta +\frac{1}{\alpha })x+(\alpha +\frac{1}{\beta })(\beta +\frac{1}{\alpha })=0$

$x^2-\left(\frac{{\alpha }^2\beta +\alpha +\alpha {\beta }^2+\beta }{\alpha \beta }\right)x+\left(\frac{{\alpha }^2{\beta }^2+1+2\alpha \beta }{\alpha \beta }\right)=0$

$x^2-\left(\frac{\alpha \beta (\alpha +\beta )+(\alpha +\beta )}{\alpha \beta }\right)x+\left(\frac{{\left(\alpha \beta \right)}^2+1+2\alpha \beta }{}\right)=0$

$x^2-\left(\frac{\frac{c}{a}\left(\frac{-b}{a}\right)+\frac{-b}{a}}{\frac{c}{a}}\right)x+\left(\frac{\frac{c^2}{a^2}+1+2\frac{c}{a}}{\frac{c}{a}}\right)=0$

$x^2-\left(\frac{\frac{-bc-ab}{a^2}}{\frac{c}{a}}\right)x+\left(\frac{\frac{c^2}{a^2}+1+2\frac{c}{a}}{\frac{c}{a}}\right)$

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

$ac{x}^2+b(a+c)x+{(c+a)}^2=0$ is the required equation.