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

21 x^{2} + 5 x - 1 = 0

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147 x^{2} + 35 x - 5 = 0

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21 x^{2} + 5 x - 3 = 0

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147 x^{2} + 35 x - 14 = 0

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Solution

The correct option is A $21 x^{2} + 5 x - 1 = 0$
Let $f (x) = 3 x^{2} + 5 x - 7 = 0$ has roots $α, β$
So, $3 α^{2} + 5 α - 7 = 0$
$\Rightarrow 3 α + 5 = \frac{7}{α}$
Now, having roots as $\frac{1}{3 α + 5}, \frac{1}{3 β + 5}$ is same as having roots $\frac{α}{7}, \frac{β}{7}$
The required equation is,
$f (7 x) = 0 \Rightarrow 3 (7 x)^{2} + 5 (7 x) - 7 = 0 ∴ 21 x^{2} + 5 x - 1 = 0$

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

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