Question

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

• A. $31$
• B. $-31$
• C. $32$
• D. $-32$

Answer 1

The correct option is B $-31$

$\Rightarrow$ $\alpha$ and $\beta$ are roots of equation $x^2-3x+4=0$

$\Rightarrow$ We get, $a=1,b=-3,c=4$

$\Rightarrow$ $\alpha \beta =\frac{c}{a}=\frac{4}{1}=4$ ----- ( 1 )

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

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

$\Rightarrow$ $(3{)}^2={\alpha }^2+{\beta }^2+2(4)$ [ From ( 1 ) and ( 2 ) ]

$\therefore$ ${\alpha }^2+{\beta }^2=9-8=1$ ------ ( 3 )

Now,

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

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

Using ( 1 ), ( 2 ) and ( 3 ) we get,

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

$\Rightarrow$ $81={\alpha }^4+{\beta }^4+16+96$

$\Rightarrow$ ${\alpha }^4+{\beta }^4=81-16-96$

$\therefore$ ${\alpha }^4+{\beta }^4=-31$

so, the correct option is '-31'.