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

x^{2} - 3 x - 2 = 0

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x^{2} - 13 x + 4 = 0

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x^{2} + 5 x + 4 = 0

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x^{2} - 5 x + 4 = 0

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Solution

The correct option is B $x^{2} - 13 x + 4 = 0$
Given equation $2 x^{2} + 3 x - 1 = 0$ has roots as $α, β$
Let $y = \frac{1}{x^{2}}$
$\Rightarrow x = \frac{1}{\sqrt{y}}$
Replace $x$ by $\frac{1}{\sqrt{y}}$ in the given equation, we get
$\frac{2}{y} + \frac{3}{\sqrt{y}} - 1 = 0 \Rightarrow (\frac{2}{y} - 1) = - \frac{3}{\sqrt{y}}$
Squaring on both sides,
${(\frac{2}{y} - 1)}^{2} = \frac{9}{y} \Rightarrow \frac{4}{y^{2}} + 1 - \frac{4}{y} = \frac{9}{y} \Rightarrow y^{2} - 13 y + 4 = 0$
So, the equation whose roots are $\frac{1}{α^{2}}, \frac{1}{β^{2}}$ is $x^{2} - 13 x + 4 = 0$

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