Question

If $\alpha ,\beta$ are the roots of $x^2-2x+4=0$, what is the value of ${\alpha }^5+{\beta }^5$?

• A. $8$
• B. $16$
• C. $32$
• D. $64$

Answer 1

The correct option is C $32$

Given, $\alpha ,\beta$ are the roots of $x^2-2x+4=0$.

Then from the relation between the roots and co-efficients we get,

$\alpha +\beta =2$.....(1) and $\alpha .\beta =4$....(2).

Now, we've,

$({\alpha }^3+{\beta }^3)({\alpha }^2+{\beta }^2)={\alpha }^5+{\beta }^5+{\alpha }^2.{\beta }^2(\alpha +\beta )$

or, $({\alpha }^3+{\beta }^3)({\alpha }^2+{\beta }^2)-{\alpha }^2.{\beta }^2(\alpha +\beta )={\alpha }^5+{\beta }^5$

or, $\left\{\right(\alpha +\beta {)}^3-3\alpha .\beta (\alpha +\beta )\left\}\right\{(\alpha +\beta {)}^2-2\alpha .\beta \}-{\alpha }^2.{\beta }^2(\alpha +\beta )={\alpha }^5+{\beta }^5$

or, ${\alpha }^5+{\beta }^5=\left\{\right(2^3-3\times 4\times 2\left)\right\}\{2^2-2\times 4\}-4^2\times 2$ [ Using (1) and (2)]

or, ${\alpha }^5+{\beta }^5=(-16)\times (-4)-32$

or, ${\alpha }^5+{\beta }^5=32$.