Question

If the roots of $x^3-12x^2+12x-28=0$ are in A.P, their common difference is

• A. $\pm 3$
• B. $\pm 2$
• C. $\pm 1$
• D. None of these

Answer 1

The correct option is A

$\pm 3$

Explanation for the correct option:

Find the value of $\mathit{a}$:

Let, $a-d,a,a+d$ be the three roots of the given cubic equation $x^3-12x^2+12x-28=0$.

Where,

$a$ is the first term and $d$ is the common difference.

Now, find $S_1$ and $S_3$. Here $S_1$ and $S_3$ are the coefficient of cubic equation.

$\begin{array}{rcl}{S}_1& =& -\frac{\mathrm{coefficient}\mathrm{of}{x}^2}{\mathrm{coefficient}\mathrm{of}{x}^3}...\left(1\right)\\ & \Rightarrow & S_1=a-d+a+a+d\\ & =& 3a\end{array}$

$\begin{array}{rcl}& \Rightarrow & 3a=-\left(-12\right)\\ a& =& \frac{12}{3}\\ & =& 4\end{array}$

Find the value of $\mathit{d}$:

$\begin{array}{rcl}{S}_3& =& \frac{-\mathrm{constant}\mathrm{term}}{\mathrm{coefficient}\mathrm{of}{x}^3}...\left(2\right)\\ & \Rightarrow & S_3=\left(a-d\right)\left(a\right)\left(a+d\right)\\ & =& \left(a^2-d^2\right)a\\ & =& \left(4^2-d^2\right)4\mathbf{[}\mathbf{\because }\mathbf{a}\mathbf{=}\mathbf{4}\mathbf{]}\\ & =& \left(16-d^2\right)4\end{array}$

$$\begin{gathered} \Rightarrow 4\left(16-d^2\right)=-\left(-28\right)\left[\mathrm{from}\left(2\right)\right] \\ \Rightarrow \left(16-d^2\right)=\frac{28}{4} \\ \Rightarrow 16-d^2=7 \\ \Rightarrow d^2=16-7 \\ \Rightarrow d^2=9 \\ \Rightarrow d=\sqrt{9} \\ \Rightarrow d=\pm 3 \end{gathered}$$

Hence, the correct option is A.