Question

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

Answer 1

Hint: We know the formula of sum of root $(\alpha + \beta )$ and product of roots $(\alpha \beta )$ hence we will try to write the desired expressions in terms of $\alpha + \beta $and $\alpha \beta $using basic algebraic identities and after writing the desired expression in terms of $\alpha + \beta $and $\alpha \beta $,we will substitute these values.

Formula Used:
1.If $a{x^2} + bx + c = 0$ is the quadratic equation having roots $\alpha $and $\beta $then
Sum of roots $(\alpha + \beta )= - \dfrac{b}{a}$
Product of roots $(\alpha \beta )$= $\dfrac{c}{a}$
2.$\left( {{a^3} + {b^3}} \right) = (a + b)({a^2} + {b^2} + ab)$

Complete step by step solution:
Given – The quadratic equation is ${x^2} + px + q = 0$
Then sum of roots $(\alpha + \beta) = - \dfrac{p}{1}$
Product of roots $(\alpha \beta) = \dfrac{q}{1}$
Let ${\alpha^3} + {\beta^3} = x$
Then using formula $\left( {{a^3} + {b^3}} \right) = (a + b)({a^2} + {b^2} + ab)$
$x = \left( {\alpha + \beta } \right)\left( {{\alpha^2} + {\beta^2} + \alpha \beta } \right)$
Simplifying the terms
$x = \left( {\alpha + \beta } \right)\left( {{\alpha^2} + {\beta^2} + 2\alpha \beta - \alpha \beta } \right)$
$ \Rightarrow x = \left( {\alpha + \beta } \right)\left( {{{\left( {\alpha + \beta } \right)}^2} - \alpha \beta } \right)$
Putting the value of $\alpha + \beta = - \dfrac{p}{1}$ and $\alpha \beta = \dfrac{q}{1}$ in $x$
$x = - p({( - p)^2} - q)$
On further simplifying the terms
$x = - p({p^2} - q)$
$ \Rightarrow x = - {p^3} + pq$
Again, let ${\alpha^4} + {\beta^4} + {\alpha^2}{\beta^2} = y$
So $y = {\alpha^4} + {\beta^4} + {\alpha^2}{\beta^2}$
Adding and subtracting ${\alpha^2}{\beta^2}$in the expression
$y = {\left( {{\alpha^2}} \right)^2} + {\left( {{\beta^2}} \right)^2} + 2{\alpha^2}{\beta^2} - {\alpha^2}{\beta^2}$
(Using ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab)$
$ \Rightarrow y = {\left( {{\alpha^2} + {\beta^2}} \right)^2} - {\alpha^2}{\beta^2}$
(Using ${a^2} + {b^2} = {\left( {a + b} \right)^2} - 2ab)$)
$ \Rightarrow y = {\left( {{{\left\{ {\alpha + \beta } \right\}}^2} - 2\alpha \beta } \right)^2} - {\alpha^2}{\beta^2}$
On expanding the square brackets
$ \Rightarrow y = {\left( {\alpha + \beta } \right)^2} + 4{\alpha^2}{\beta^2} - 2\alpha \beta \left( {\alpha + \beta } \right) - {\alpha^2}{\beta^2}$
$ \Rightarrow y = {\left( {\alpha + \beta } \right)^2} - 2\alpha \beta \left( {\alpha + \beta } \right) + 3{\alpha^2}{\beta^2}$
Putting the value of $\alpha + \beta = - \dfrac{p}{1}$ and $\alpha \beta =\dfrac{q}{1}$ in $y$
$y = {( - p)^2} + 3{(q)^2} - 2q( - p)$
On simplifying the terms
$ \Rightarrow y = {p^2} + 3{q^2} + 2pq$
Rearranging the terms in expression
$ \Rightarrow y = {p^2} + 2pq + 3{q^2}$

Hence value of ${\alpha^3} + {\beta^3}$is $ - {p^3} + pq$ and ${\alpha^4} + {\beta^4} + {\alpha^2}{\beta^2}$ is ${p^2} + 2pq + 3{q^2}$.

Note: In such types of questions, always try to write the asked expression in terms of $\alpha + \beta $and $\alpha \beta $.
Following formulas are useful in simplifying the expression –
$\left( {{a^3} + {b^3}} \right) = (a + b)({a^2} + {b^2} + ab)$
${a^4} + {b^4} = {\left( {a + b} \right)^2} + 2{\left( {ab} \right)^2} - 2ab\left( {a + b} \right)$