Find the equation whose roots are the squares of the roots of $x^{4} + x^{3} + 2 x^{2} + x + 1 = 0$ .

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Solution

$x^{4} + x^{3} + 2 x^{2} + x + 1 = 0$

$y = x^{2} ⟹ x = \sqrt{y} ∴$ replace $x$ with $\sqrt{y}$ in the given question

$⟹ (\sqrt{y})^{4} + (\sqrt{y})^{3} + 2 (\sqrt{y})^{2} + \sqrt{y} + 1 = 0$

$⟹ y^{2} + y \sqrt{y} + 2 y + \sqrt{y} + 1 = 0$

$⟹ \sqrt{y} (y + 1) = - (y^{2} + 2 y + 1)$

$⟹ y (y + 1)^{2} = (y^{2} + 2 y + 1)^{2} [taking squares on both sides to remove the radical sign]$

$⟹ y (y^{2} + 2 y + 1) = (y^{4} + 4 y^{3} + 6 y^{2} + 4 y + 1)$

$⟹ y^{3} + 2 y^{2} + y = y^{4} + 4 y^{3} + 6 y^{2} + 4 y + 1$

$⟹ y^{4} + 3 y^{3} + 4 y^{2} + 3 y + 1 = 0$

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Find the equation whose roots are the squares of the roots of x4+x3+2x2+x+1=0.

Find the equation whose roots are the squares of the roots of $x^{4} + x^{3} + 2 x^{2} + x + 1 = 0$ .