Question

If $\alpha$and $\beta$are the roots of the quadratic equation$x^2+px+q=0$, then the values of ${\alpha }^3+{\beta }^3$ and ${\alpha }^4+{\alpha }^2{\beta }^2+{\beta }^4$ are respectively

• A. $3pq–p^3$and $p^4–3p^{2}q+3q^2$
• B. $-p(3q–p^2)$and $(p^2–q)(p^2+3q)$
• C. $pq–4$and $p^4–q^4$
• D. $3pq–p^3$ and $(p^2–q)(p^2–3q)$

Answer 1

The correct option is D

$3pq–p^3$ and $(p^2–q)(p^2–3q)$

Explanation for correct option:

Find the values of ${\mathit{\alpha }}^{\mathbf{3}}\mathbf{+}\mathbf{}{\mathit{\beta }}^{\mathbf{3}}$ and ${\mathit{\alpha }}^{\mathbf{4}}\mathbf{+}\mathbf{}{\mathit{\alpha }}^{\mathbf{2}}{\mathit{\beta }}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}{\mathit{\beta }}^{\mathbf{4}}$:

Given that If $\alpha$and $\beta$are the roots of the quadratic equation$x^2+px+q=0$

$\Rightarrow (\alpha +\beta )=-p;\alpha \beta =q$

Now, ${\alpha }^3+{\beta }^3={(ɑ+\beta )}^3–3ɑ\beta [ɑ+\beta ]$

$$\begin{gathered} ={(-p)}^3–3q(-p) \\ =–p^3+3pq \end{gathered}$$

${\alpha }^3+{\beta }^3={-p}^3+3pq$

And ${\alpha }^4+{\alpha }^2{\beta }^2+{\beta }^4={({\alpha }^2+{\beta }^2)}^2–(ɑ\beta {)}^2$

$$\begin{gathered} ={[{(ɑ+\beta )}^2–2ɑ\beta ]}^2–(ɑ\beta {)}^2={[{(-p)}^2–2q]}^2–q^2 \\ ={(p^2–2q)}^2–q^2=p^4+4q^2–4p^{2}q–q^2 \\ =p^4+3q^2–4p^{2}q \\ =(p^2–q)(p^2–3q) \end{gathered}$$

${\alpha }^4+{\alpha }^2{\beta }^2+{\beta }^4=(p^2–q)(p^2-3q)$

Hence, the correct option is (D).