Question

# If roots of cubic equation $a{{x}^{3}}+b{{x}^{2}}+cx+d$ are in G.P., then:\begin{align} & A.\text{ }{{c}^{3}}a={{b}^{3}}d \\ & B.\text{ }{{a}^{2}}c={{b}^{2}}d \\ & C.\text{ }a{{c}^{2}}=b{{d}^{2}} \\ & D.\text{ NOTA} \\ \end{align}

Hint: At first, take roots of the equation as $\alpha ,\alpha r,\alpha {{r}^{2}}$ with r as common ratio, then, use the relation between the roots and coefficient to get relation between a, b, c, d. Sum of roots is related to the ratio of –b and a of the quadratic equation whereas the product of roots is given by ratio of -d and a. Then we have another relation which gives the sum of product of two consecutive roots as ratio of c and a.

In the question, if the roots of cubic equation $a{{x}^{3}}+b{{x}^{2}}+cx+d$ are in G.P, then, we have to find a relation between a, b, c, d.
Now, as we know that, the roots are in G.P. So, let's suppose the roots are $\alpha ,\alpha r,\alpha {{r}^{2}}$ with a common ratio 'r'.
So, by relation between the roots and the coefficient of equation, we can say that,
Sum of the roots $\Rightarrow -\dfrac{b}{a}$
$\Rightarrow \alpha +\alpha r+\alpha {{r}^{2}}=-\dfrac{b}{a}$
Which can be written as,
$\alpha \left( 1+r+{{r}^{2}} \right)=-\dfrac{b}{a}$
Product of the roots $\Rightarrow \dfrac{c}{a}$
\begin{align} & \Rightarrow \alpha \times \alpha r+\alpha \times \alpha {{r}^{2}}+\alpha r\times \alpha {{r}^{2}}=\dfrac{c}{a} \\ & \Rightarrow {{\alpha }^{2}}r+{{\alpha }^{2}}{{r}^{2}}+{{\alpha }^{2}}{{r}^{3}}=\dfrac{c}{a} \\ \end{align}
Which can be written as,
${{\alpha }^{2}}r\left( 1+r+{{r}^{2}} \right)=\dfrac{c}{a}$
We can written it as,
$\alpha r\times \alpha \left( 1+r+{{r}^{2}} \right)=\dfrac{c}{a}$
Now, we can write $\alpha \left( 1+r+{{r}^{2}} \right)\text{ as -}\dfrac{b}{a}$ as we know that $\alpha \left( 1+r+{{r}^{2}} \right)\text{ as -}\dfrac{b}{a}$
\begin{align} & \alpha r\times -\dfrac{b}{a}=\dfrac{c}{a} \\ & \Rightarrow \alpha r=-\dfrac{c}{b} \\ \end{align}
The final relation between roots and coefficient of equation is,
\begin{align} & \alpha \times \alpha r\times \alpha {{r}^{2}}=-\dfrac{d}{a} \\ & \Rightarrow {{\alpha }^{3}}{{r}^{3}}=-\dfrac{d}{a} \\ \end{align}
We know that, $\alpha r=-\dfrac{c}{b}$
So, we can write,
${{\alpha }^{3}}{{r}^{3}}={{\left( \alpha r \right)}^{3}}\Rightarrow {{\left( -\dfrac{c}{b} \right)}^{3}}$
We found that ${{\alpha }^{3}}{{r}^{3}}\text{ is }-\dfrac{d}{a}$ so, we can write,
\begin{align} & {{\left( -\dfrac{c}{b} \right)}^{3}}=-\dfrac{d}{a} \\ & \Rightarrow -\dfrac{{{c}^{3}}}{{{b}^{3}}}=-\dfrac{d}{a} \\ \end{align}
Now, on cross multiplication we get,
${{c}^{3}}a={{b}^{3}}d$
So, the correct answer is “Option A”.

Note: When it is said that, roots are given in G.P, then we can take the value of three roots as any three consecutive terms of any G.P such as $$\alpha ,\alpha r,\alpha {r^2}$$ by reducing the three variables into two. Also, while taking this, one should know here 'r' which is a common ratio cannot be equal to '0'. If 'r' is equal to '0' then, it will not satisfy the condition of being in G.P.