Question

If $\alpha, \beta$ are the roots of the equation $3x^2 +5x -7 = 0$, then the equation whose roots are $\dfrac{1}{3\alpha+5}, \dfrac{1}{3\beta+5}$ is

Answer 1

Let $f(x) = 3x^2 +5x-7 = 0$ has roots $\alpha, \beta$
So, $3\alpha^2 +5\alpha -7 = 0$
$\Rightarrow 3\alpha+5 = \dfrac{7}{\alpha}$
Now, having roots as $\dfrac{1}{3\alpha+5}, \dfrac{1}{3\beta+5}$ is same as having roots $\dfrac{\alpha}{7}, \dfrac \beta 7 $
The required equation is,
$f(7x) = 0\\
\Rightarrow 3(7x)^2 +5(7x) -7 = 0\\
\therefore 21x^2 +5x-1 = 0$

Alternate Solution :
$\dfrac{1}{3x+5}=y$
$\Rightarrow x=\dfrac{1-5y}{3y}$
Therefore, equation whose roots are $\dfrac{1}{3\alpha+5}, \dfrac{1}{3\beta+5}$ is
$f\left(\dfrac{1-5x}{3x}\right)=0$
$\Rightarrow 3\left(\dfrac{1-5x}{3x}\right)^2 +5\left(\dfrac{1-5x}{3x}\right)-7 = 0$
$\Rightarrow 21x^2 +5x-1 = 0$