જો alpha અને beta એ સમીકરણ x^{2}+(3)^{1 / 4} | vedclass

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Question

As, $\left(a^{2}+\sqrt{3}ight)=-(3)^{1 / 4} \cdot \alpha$

$\Rightarrow\left(\alpha^{2}+2 \sqrt{3} \alpha^{2}+3ight)=\sqrt{3} \alpha^{2} 	ext {

Vedclass · Accepted Answer

$52 	imes 3^{24}$

Vedclass · Answer

As, $\left(a^{2}+\sqrt{3}ight)=-(3)^{1 / 4} \cdot \alpha$

$\Rightarrow\left(\alpha^{2}+2 \sqrt{3} \alpha^{2}+3ight)=\sqrt{3} \alpha^{2} 	ext { (On squaring) }$

$\left.	herefore\left(a^{4}+3ight)=(-) \sqrt{3} \alpha^{2}ight)$

$\Rightarrow \alpha^{8}+6 \alpha^{4}+9=3 \alpha^{2}$ (Again squaring)

$	herefore \alpha^{8}+3 \alpha^{4}+9=0$

$\Rightarrow \alpha^{8}=-9-3 \alpha^{4}$

(Multiply by $\left.\alpha^{4}ight)$

So, $\alpha^{12}=-9 \alpha^{4}-3 \alpha^{8}$

$	herefore \alpha^{12}=-9 \alpha^{4}-3\left(-9-3 \alpha^{4}ight)$

$\Rightarrow \alpha^{12}=-9 a^{4}+27+9 a^{4}$

Hence, $\alpha^{12}=(27)$

$\Rightarrow\left(\alpha^{12}ight)^{18}=(27)^{8}$

$\Rightarrow \alpha^{96}=(3)^{24}$

Similarly $\beta^{96}=(3)^{24}$

$	herefore \alpha^{96}\left(\alpha^{12}-1ight)+\beta^{96}\left(\beta^{12}-1ight)=(3)^{24} 	imes 52$

જો $\alpha$ અને $\beta$ એ સમીકરણ $x^{2}+(3)^{1 / 4} x+3^{1 / 2}=0$ નાં ભિન્ન બીજ હોય તો $\alpha^{96}\left(\alpha^{12}-\right.1) +\beta^{96}\left(\beta^{12}-1\right)$ ની કિમંત મેળવો.

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