The equation whose roots are the squares of the roots of $x^{4} + 3 x + 8 = 0$ , is:

x^{4} - 16 x^{2} - 9 x - 64 = 0

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x^{4} + 16 x^{2} - 9 x - 64 = 0

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x^{4} + 16 x^{2} - 9 x + 64 = 0

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x^{4} - 16 x^{2} + 9 x - 64 = 0

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Solution

The correct option is B $x^{4} + 16 x^{2} - 9 x + 64 = 0$
Let $y = x^{2}$ ...(1)
$x^{4} + 3 x + 8 = 0 \Rightarrow x^{4} + 8 = - 3 x \Rightarrow {(x^{4} + 8)}^{2} = {(- 3 x)}^{2} \Rightarrow x^{8} + 64 + 16 x^{4} = 9 x^{2} \Rightarrow x^{8} + 16 x^{4} - 9 x^{2} + 64 = 0$

Substituting value from (1), we have
$y^{4} + 16 y^{2} - 9 y + 64 = 0$
Replacing $x$ by $y$ , we get
$x^{4} + 16 x^{2} - 9 x + 64 = 0$

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