JEE Main 2024 — Complex Number Question with Solution

Given equation $x^2-x+2$ has roots $\alpha \& \beta$,

So, $\alpha +\beta =1, \alpha \beta =2$

$\Rightarrow {\alpha }^4+{\beta }^4={\left({\alpha }^2+{\beta }^2\right)}^2-2{\alpha }^2{\beta }^2$

$\Rightarrow {\alpha }^4+{\beta }^4={\left[{\left(\alpha +\beta \right)}^2-2\alpha \beta \right]}^2-2{\alpha }^2{\beta }^2$

$\Rightarrow {\alpha }^4+{\beta }^4={\left(1-4\right)}^2-2\times 4$

$\Rightarrow {\alpha }^4+{\beta }^4=1$

It is given that, $\alpha$ is a root of $x^2-x+2$.

$\Rightarrow {\alpha }^2-\alpha +2=0$

$\Rightarrow {\alpha }^2=\alpha -2$

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

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

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

Now, ${\alpha }^6-5{\alpha }^2={\alpha }^2\left({\alpha }^4-5\right)$

$\Rightarrow {\alpha }^6-5{\alpha }^2=\left(\alpha -2\right)\left(2-3\alpha -5\right)$

$\Rightarrow {\alpha }^6-5{\alpha }^2=\left(\alpha -2\right)\left(-3\alpha -3\right)$

$\Rightarrow {\alpha }^6-5{\alpha }^2=-3\left({\alpha }^2-\alpha -2\right)$

$\Rightarrow {\alpha }^6-5{\alpha }^2=-3\left(\alpha -2-\alpha -2\right)$

$\Rightarrow {\alpha }^6-5{\alpha }^2=12$

$\Rightarrow {\alpha }^6+{\alpha }^4+{\beta }^4-5{\alpha }^2=13$