Question

If $\alpha ,\beta ,\gamma$ are the roots of $x^3+p{x}^2+qx+r=0$, then $\Sigma {\alpha }^2{\beta }^2=$

• A. $q^2-2pr$
• B. $q^2+2pr$
• C. $q+2pr$
• D. $q-2pr$

Answer 1

Answer: (A)

$q^2-2pr$

Since $\alpha ,\beta ,$ and $\gamma$ are the roots of given cubic equation

Therefore $(x-\alpha )(x-\beta )(x-\gamma )=0$

By comparing this with given cubic,

we got $\alpha \beta \gamma =-r$..........(1)

$\alpha \beta +\beta \gamma +\gamma \alpha =q$.....(2)

$-(\alpha +\beta +\gamma )=p$.....(3)

Now to get $\sum {\alpha }^2{\beta }^2$, we add two times the product of Eq(1) and Eq(3) to square of Eq(2)

we got $\sum {\alpha }^2{\beta }^2=q^2-2pr$

Hence, option A is correct.