Let $α, β$ be the roots of the quadratic equation $3 x^{2} + 10 x + 2 = 0$ , then the quadratic equation whose roots are $\frac{α}{α + 5}, \frac{β}{β + 5}$ , is

27 x^{2} + 48 x + 2 = 0

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27 x^{2} + 46 x + 2 = 0

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46 x^{2} + 27 x + 2 = 0

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27 x^{2} + 2 x + 48 = 0

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Solution

The correct option is B $27 x^{2} + 46 x + 2 = 0$
Let $y = \frac{x}{x + 5}$ , where $x = α, β$
$x y + 5 y = x \Rightarrow x = \frac{5 y}{1 - y}$
Putting $x = \frac{5 y}{1 - y}$ in the quadratic equation
$3 x^{2} + 10 x + 2 = 0 \Rightarrow 3 {(\frac{5 y}{1 - y})}^{2} + 10 (\frac{5 y}{1 - y}) + 2 = 0 \Rightarrow 75 y^{2} + 50 y (1 - y) + 2 (1 - y)^{2} = 0 \Rightarrow 27 y^{2} + 46 y + 2 = 0$

Hence, the quadratic equation whose roots are $\frac{α}{α + 5}, \frac{β}{β + 5}$ is
$27 x^{2} + 46 x + 2 = 0$

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