Question

If roots of the cubic equation $64x^3-144x^2+92x-15=0$ are in arithmetic progression, then the difference between the largest and smallest root is equal to

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{1}{4},\frac{3}{4},\frac{5}{4}$$\frac{7}{8}$
• D. None of these

Answer 1

The correct option is A

$1$

Explanation for the correct option:

Step $1:$ Find the roots of the given cubic equation.

If a cubic equation is in the form of $a{x}^3+b{x}^2+cx+d=0$.

Then, the product of roots of the cubic equation can be given by $-\frac{b}{a}$ and the sum of roots of the cubic equation can be given by $-\frac{d}{a}$.

Where, $a,b$ and $d$ are the coefficients of $x^3,x^2$ and the constant term respectively.

Assume that, $p-q,p,p+q$ be the roots of the given cubic equation.

Therefore, $p-q+p+p+q=\frac{144}{64}$

$$\begin{gathered} \Rightarrow 3p=\frac{9}{4} \\ \Rightarrow p=\frac{3}{4} \end{gathered}$$

Also, $\left(p-q\right)p\left(p+q\right)=\frac{15}{64}$

$$\begin{gathered} \Rightarrow \left(p^2-q^2\right)p=\frac{15}{64} \\ \Rightarrow p^2-q^2=\frac{15}{64p} \\ \Rightarrow p^2-\frac{15}{64p}=q^2 \\ \Rightarrow q=\sqrt{p^2-\frac{15}{64p}} \end{gathered}$$

Since, $p=\frac{3}{4}$.

$$\begin{gathered} \Rightarrow q=\sqrt{{\left(\frac{3}{4}\right)}^2-\frac{15}{64\times {\displaystyle \frac{3}{4}}}} \\ \Rightarrow q=\sqrt{\frac{9}{16}-\frac{5}{16}} \\ \Rightarrow q=\sqrt{\frac{4}{16}} \\ \Rightarrow q=\sqrt{\frac{1}{4}} \end{gathered}$$

Therefore, $q=\frac{1}{2}$.

Also, the value of $p-q,p,p+q$ are $\frac{1}{4},\frac{3}{4},\frac{5}{4}$ respectively.

Step $2:$ Calculate the required difference.

The difference between the largest and the smallest root is $p+q-\left(p-q\right)=\frac{5}{4}-\frac{1}{4}$

$$\begin{gathered} \Rightarrow p+q-\left(p-q\right)=\frac{4}{4} \\ \Rightarrow p+q-\left(p-q\right)=1 \end{gathered}$$

Hence, option $A$ is the correct answer.