Mathematics - Quadratic Equation Question with Solution | TestHub

We have,

$x^2+{\left(3\right)}^{\frac{1}{4}}x+3^{\frac{1}{2}}=0$

Since, $\alpha$ is the root, therefore $\left({\alpha }^2+\sqrt{3}\right)=-{\left(3\right)}^{\frac{1}{4}}·\alpha$

On squaring, both sides we get

$\left({\alpha }^4+2\sqrt{3}{\alpha }^2+3\right)=\sqrt{3}{\alpha }^2$

$\Rightarrow {\alpha }^4+3=-\sqrt{3}{\alpha }^2$

On squaring, both sides we get

${\alpha }^8+6{\alpha }^4+9=3{\alpha }^4$

$\Rightarrow {\alpha }^8=-9-3{\alpha }^4$

Multiply by ${\alpha }^4$, we get

${\alpha }^{12}=-9{\alpha }^4-3{\alpha }^8$

$\therefore {\alpha }^{12}=-9{\alpha }^4-3\left(-9-3{\alpha }^4\right)$

$\Rightarrow {\alpha }^{12}=27$

$\Rightarrow {\left({\alpha }^{12}\right)}^8={\left(27\right)}^8$

$\Rightarrow {\alpha }^{96}={\left(3\right)}^{24}$

Similarly,

${\beta }^{96}={\left(3\right)}^{24}$

$\therefore {\alpha }^{96}\left({\alpha }^{12}-1\right)+{\beta }^{96}\left({\beta }^{12}-1\right)=\left(3\right){}^{24}+\left(27+27-2\right)={\left(3\right)}^{24}\times 52$