Question

If α,β are the roots of the quadratic equation px2 + qx + r = 0 the α2 /β + β2 /α =​

Answer 1

GIVEN :–

• A quadratic equation px² + qx + r = 0.

• It has two roots α and β.

TO FIND :–

• Value of α²/β + β²/α = ?

SOLUTION :–

• We know that –

$\\ \implies \bf Sum \: \: of \: \: roots = - \dfrac{Coefficient \: \:of \: \: x}{Coefficient \: \:of \: \: {x}^{2} }$

$\\ \implies \bf \alpha + \beta = - \dfrac{q}{p} \: \: \: \: \: \: \: \: \: \: \: - - - - eq.(1)$

• And –

$\\ \implies \bf Product \: \: of \: \: roots = - \dfrac{Constant \: \: term}{Coefficient \: \:of \: \: {x}^{2} }$

$\\ \implies \bf \alpha\beta = \dfrac{r}{p} \: \: \: \: \: \: \: \: \: \: \: - - - - eq.(2)$

• Now Let's find α²/β + β²/α :–

• Let –

$\\ \implies \bf A = \dfrac{ { \alpha }^{2} }{ \beta } + \dfrac{ { \beta }^{2} }{ \alpha }$

$\\ \implies \bf A = \dfrac{ { \alpha }^{3} + { \beta }^{3}}{ \alpha\beta }$

• Using identity –

$\\ \implies \large{ \boxed{ \bf {a}^{3} + {b}^{3} = {(a + b)}^{3} - 3ab(a + b)}}$

• So that –

$\\ \implies \bf A = \dfrac{ { (\alpha + \beta ) }^{3} - 3 \alpha \beta ( \alpha + \beta )}{ \alpha\beta }$

• Now put the values –

$\\ \implies \bf A = \dfrac{ { \bigg(- \dfrac{q}{p} \bigg) }^{3} - 3 \bigg(\dfrac{r}{p} \bigg) \bigg(- \dfrac{q}{p}\bigg)}{\bigg(\dfrac{r}{p} \bigg)}$

$\\ \implies \bf A = \dfrac{ { - \bigg( \dfrac{q}{p} \bigg) }^{3} + 3 \bigg(\dfrac{qr}{p^{2} } \bigg) }{\bigg(\dfrac{r}{p} \bigg)}$

$\\ \implies \bf A = \dfrac{ { - \bigg( \dfrac{q}{p} \bigg) }^{3} + 3 \bigg(\dfrac{pqr}{p^{3} } \bigg) }{\bigg(\dfrac{r}{p} \bigg)}$

$\\ \implies \bf A = \dfrac{ \bigg(\dfrac{3pqr - {q}^{3} }{p^{3} } \bigg) }{\bigg(\dfrac{r}{p} \bigg)}$

$\\ \implies \bf A = \bigg(\dfrac{3pqr - {q}^{3} }{p^{3} } \bigg)\bigg(\dfrac{p}{r} \bigg)$

$\\ \implies \large { \boxed{\bf A = \bigg(\dfrac{3pqr - {q}^{3} }{ {p}^{2} r} \bigg)}}$

• Hence –

$\\ \implies \large { \boxed{\bf \dfrac{ { \alpha }^{2} }{ \beta } + \dfrac{ { \beta }^{2} }{ \alpha } = \bigg(\dfrac{3pqr - {q}^{3} }{ {p}^{2} r} \bigg)}}$