Question

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

• A. $147x^2+35x-5=0$
• B. $147x^2+35x-14=0$
• C. $21x^2+5x-1=0$
• D. $21x^2+5x-3=0$

Answer 1

The correct option is C $21x^2+5x-1=0$

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

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