Question

Which of the option(s) is/are correct “If one root of \[a{x^2} + bx + c = 0\] be the square of the other, then the value of ${b^3} + a{c^2} + {a^2}c$ is:”
A) $3abc$
B) $ - 3abc$
C) $0$
D) None of these

Answer 1

Hint:Let $a{x^2} + bx + c = 0$ where a,b and c are constants and $a \ne 0$ let $\alpha $and $\beta $are the roots of this equation then : Sum of the roots = $\alpha + \beta = - \dfrac{b}{a}$ and Product of the roots= $\alpha \beta = \dfrac{c}{a}$.Consider the equation $a{x^3} + b{x^2} + cx + d = 0$where a,b,c and d are constants and $a \ne 0$ and let $\alpha ,\beta $and $\gamma $are the roots of this equation then: Sum of the roots= $\alpha + \beta + \gamma = \dfrac{{ - b}}{a}$ and Product of the roots= $\alpha \beta \gamma = \dfrac{{ - d}}{a}$.Identity: ${(a + b)^3} = {a^3} + {b^3} + 3ab(a + b)$

Complete step-by-step answer:
According to the question, one root of a quadratic polynomial \[a{x^2} + bx + c\]is square of the other.
So let’s assume its roots are $\alpha $ and ${\alpha ^2}$.
We have to find the value of ${b^3} + a{c^2} + {a^2}c$.
We know that, if and $\beta $are the roots of the quadratic equation $a{x^2} + bx + c = 0$then
Sum of the roots of quadratic equation = $\alpha + \beta = - \dfrac{b}{a}$ and Product of the roots of quadratic equation=$\alpha \beta = \dfrac{c}{a}$.
Here we have roots of the quadratic equation \[a{x^2} + bx + c = 0\]as$\alpha $and${\alpha ^2}$.
$\therefore $The sum of roots of quadratic equation = $\alpha + {\alpha ^2} = - \dfrac{b}{a}$ ……………………….(i) and,
The product of the roots of quadratic equation=$\alpha \times {\alpha ^2} = \dfrac{c}{a}$.
                                                                                      $ \Rightarrow {\alpha ^3} = \dfrac{c}{a}$ ………………………….(ii)
We have to find the value of expression in which we have ${b^3}$.
By cubing equation (i), we get
${(\alpha + {\alpha ^2})^3} = {\left( {\dfrac{{ - b}}{a}} \right)^3}$
By applying the identity: ${(a + b)^3} = {a^3} + {b^3} + 3ab(a + b)$.
Taking $a = \alpha $and$b = {\alpha ^2}$we get,
${\alpha ^3} + {({\alpha ^2})^3} + 3 \times \alpha \times {\alpha ^2}(\alpha + {\alpha ^2}) = \dfrac{{{{( - b)}^3}}}{{{a^3}}}$
By using property of exponents: ${({a^n})^m} = {a^{nm}}$, we get
$ \Rightarrow {\alpha ^3} + {\alpha ^6} + 3{\alpha ^3}(\alpha + {\alpha ^2}) = \dfrac{{{{( - b)}^3}}}{{{a^3}}}$
Again by using property of exponents: ${({a^n})^m} = {a^{nm}}$, we get
$ \Rightarrow {\alpha ^3} + {\alpha ^6} + 3{\alpha ^4} + 3{\alpha ^5} = \dfrac{{ - {b^3}}}{{{a^3}}}$
Now we convert the above equation in the form of $(\alpha + {\alpha ^2})$and ${\alpha ^3}$form.
$ = {\alpha ^3}(1 + {\alpha ^3} + 3\alpha + 3{\alpha ^2}) = \dfrac{{ - {b^3}}}{{{a^3}}}$
By taking 3 common from$3\alpha + 3{\alpha ^2}$, we get
${\alpha ^3}[1 + {\alpha ^3} + 3(\alpha + {\alpha ^2})] = \dfrac{{ - {b^3}}}{{{a^3}}}$ ………………………(iii)
So now substitute the values in the above equation from equation (i) and (ii).
We have, $\dfrac{c}{a}\left\{ {1 + \dfrac{c}{a} + 3\left( {\dfrac{{ - b}}{a}} \right)} \right\} = \dfrac{{ - {b^3}}}{{{a^3}}}$
Solving the bracket first, we get
$ \Rightarrow \dfrac{c}{a} + \dfrac{{{c^2}}}{{{a^2}}} - \dfrac{{3bc}}{{{a^2}}} = \dfrac{{ - {b^3}}}{{{a^3}}}$
By taking ${a^2}$ as L.C.M
$ \Rightarrow \left( {\dfrac{{ac + {c^2} - 3bc}}{{{a^2}}}} \right) = \dfrac{{ - {b^3}}}{{{a^3}}}$
Now taking ${a^3}$on the L.H.S, we get
$ \Rightarrow {a^2}c + a{c^2} - 3abc = - {b^3}$
Or, ${b^3} + {a^2}c + a{c^2} = 3abc$

So, the correct answer is “Option A”.

Note:Alternative Method:
We can also solve the above question by simply substituting the values of b and c in ${b^3} + a{c^2} + {a^2}c$ as
The sum of roots of quadratic equation = $\alpha + {\alpha ^2} = - \dfrac{b}{a}$
$\therefore b = - a\alpha (1 + \alpha )$
The product of the roots of quadratic equation=$\alpha \times {\alpha ^2} = \dfrac{c}{a}$.
 $ \Rightarrow a{\alpha ^3} = c$
Now consider,${b^3} + a{c^2} + {a^2}c$
$ = {( - a\alpha (1 + \alpha ))^3} + {a^2} \times a{\alpha ^3} + a \times {(a{\alpha ^3})^2}$
Solving the bracket first, we get
$ \Rightarrow - {a^3}{\alpha ^3}{(1 + \alpha )^3} + {a^2} \times a{\alpha ^3} + a \times ({a^2}{\alpha ^6})$
Taking ${a^3}{m^3}$and by applying the identity: ${(a + b)^3} = {a^3} + {b^3} + 3ab(a + b)$.
$ = - 3{a^3}{\alpha ^3}(1 + \alpha )\alpha $
$ = - 3a \times a{\alpha ^3} \times a(\alpha + {\alpha ^2})$
As the value of $b = - a\alpha (1 + \alpha )$
$ \Rightarrow - 3ac( - b)$
$ \Rightarrow 3abc$
Or, ${b^3} + {a^2}c + a{c^2} = 3abc$
$\therefore $The Option A is the correct answer.