Question

If $\alpha ,\beta$ are the roots of x${}^2$-x+3=0 then ${\alpha }^4+{\beta }^4=$

• A. 10
• B. 7
• C. -10
• D. 8

Answer 1

The correct option is B 7

$\Rightarrow$ $\alpha$ and $\beta$ are roots of the equation $x^2-x+3=0$

Here, $a=1,b=-1,c=3$

$\Rightarrow$ $\alpha +\beta =\frac{-b}{a}=\frac{-(-1)}{1}=1$ ----- ( 1 )

$\Rightarrow$ $\alpha \beta =\frac{c}{a}=\frac{3}{1}=3$ ---- ( 2 )

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

$\Rightarrow$ $(1{)}^2={\alpha }^2+{\beta }^2+2(3)$ [ From ( 1 ) and ( 2 ) ]

$\Rightarrow$ $1={\alpha }^2+{\beta }^2+6$

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

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

$\Rightarrow$ $(1{)}^2={\alpha }^4+{\beta }^4+4\alpha \beta ({\alpha }^2+{\beta }^2)+6{\alpha }^2{\beta }^2$

$\Rightarrow$ $1={\alpha }^4+{\beta }^4+4\left(3\right)(-5)+6(3{)}^2$

$\Rightarrow$ $1={\alpha }^4+{\beta }^4-60+54$

$\Rightarrow$ $1={\alpha }^4+{\beta }^4-6$

$\therefore$ ${\alpha }^4+{\beta }^4=7$