Question

If $\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$ then the value of ${\alpha }^6+{\beta }^6$ is equal to

• A. $32$
• B. $64$
• C. $128$
• D. $256$

Answer 1

The correct option is A $32$

Given equation $\Rightarrow x^2-2x+4=0$

Find ${\alpha }^6+{\beta }^6$

Considering $\Rightarrow x^2-2x+4=0$

$\alpha +\beta =\frac{-b}{a}=2$

$\alpha \beta =\frac{c}{a}=4$

$\Rightarrow {\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta =4-8=-4$

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

$({\alpha }^2+{\beta }^2{)}^2=16$

$\Rightarrow {\alpha }^4+{\beta }^4+2{\alpha }^2{\beta }^2=16$

$\Rightarrow {\alpha }^4+{\beta }^4=16-2{\alpha }^2{\beta }^2=-16$

$(\alpha {)}^6+(\beta {)}^6=({\alpha }^2{)}^3+({\beta }^2{)}^3=({\alpha }^2+{\beta }^2)({\alpha }^4+{\beta }^4-{\alpha }^2{\beta }^2)$

$=(-4)(-16-16)$

$\Rightarrow (-4)(-32)$

$\therefore (\alpha {)}^6+(\beta {)}^6=128$.