alpha=sin 36^{circ} is a root of which of the | vedclass

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Question

$\cos 72^{\circ}=\frac{\sqrt{5}-1}{4}$

$\Rightarrow 1-2 \sin ^{2} 36^{\circ}=\frac{\sqrt{5}-1}{4}$

$\Rightarrow 4-8 \alpha^{2}=\sqrt{5}-1$4

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Vedclass · Accepted Answer

$16 x^{4}-20 x^{2}+5=0$

Vedclass · Answer

$\cos 72^{\circ}=\frac{\sqrt{5}-1}{4}$

$\Rightarrow 1-2 \sin ^{2} 36^{\circ}=\frac{\sqrt{5}-1}{4}$

$\Rightarrow 4-8 \alpha^{2}=\sqrt{5}-1$4

$\Rightarrow 5-8 \alpha^{2}=\sqrt{5}$

$\Rightarrow\left(5-8 \alpha^{2}\right)^{2}=5$

$\Rightarrow 25+64 \alpha^{4}-80 \alpha^{2}=5$

$\Rightarrow 64 \alpha^{4}-80 \alpha^{2}+20=0$

$\Rightarrow 16 \alpha^{4}-20 \alpha^{2}+5=0$

$\alpha=\sin 36^{\circ}$ is a root of which of the following equation

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