Question

If $\alpha ,\beta$ are the roots of the equation $x^2-px+q=0$, find the value of ${\alpha }^4+{\alpha }^2{\beta }^2+{\beta }^4$.

Answer 1

$\Rightarrow$ $\alpha$ and $beta$ are the roots of the equestion, $x^2-px+q=0$

$\Rightarrow$ Here, $a=1,b=-p,c=q$

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

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

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

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

$\Rightarrow$ $(p{)}^2={\alpha }^2+{\beta }^2+2(q)$ [ By using ( 1 ) and ( 3 ) ]

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

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

$\Rightarrow$ $(p{)}^4={\alpha }^4+{\beta }^4+6(q{)}^2+4\left(q\right)(p^2-2q{)}^2$ ------ [ By using ( 1 ), ( 2 ) and ( 4 ) ]

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

$\Rightarrow$ $p^4={\alpha }^4+{\beta }^4+6q^2+4q{p}^4-16p^2{q}^2+16q^3$

$\Rightarrow$ ${\alpha }^4+{\beta }^4=p^4-6q^2-4q{p}^4+16p^2{q}^2-16q^3$

$\Rightarrow$ ${\alpha }^4+{\beta }^4=p^4(1-4q)+16q^2(p^2-q)-6q^2$ ----- ( 5 )

Now,

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

$=p^4(1-4q)+16q^2(p^2-q)-6q^2+q^2$

$\therefore$ ${\alpha }^4+{\alpha }^2{\beta }^2+{\beta }^4=p^4(1-4q)+16q^2(p^2-q)-5q^2$