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

Answer: (B)

$7$

$x^2-x+3=0$

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

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

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

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

$\Rightarrow$ $=(-5{)}^2-2(3{)}^2=7$