Question

If roots of the equation $x^3-12x^2+39x-28=0$ are in A.P., then its common difference is -

• A. $\pm 1$
• B. $\pm 2$
• C. $\pm 3$
• D. $\pm 4$

Answer 1

The correct option is C $\pm 3$

Let the roots of the given equation $x^3-12x^2+39x-28=0$ be $a-d,a$ and $a+d$ (Given that the roots are in A.P).

We know that the sum of the roots of a quadratic equation $a{x}^3+b{x}^2+cx+d=0$ is $-\frac{b}{a}$ and the product of the roots is $\frac{d}{a}$.

Here, the equation is $x^3-12x^2+39x-28=0$, therefore, we have:

Sum of the roots is:

$(a-d)+a+(a+d)=-\frac{(-12)}{1}\Rightarrow 3a=12\Rightarrow a=\frac{12}{3}\Rightarrow a=4.....\left(1\right)$

Product of the roots is:

$(a-d)a(a+d)=28\Rightarrow (4-d)4(4+d)=28\left(Fromeqn\right(1\left)\right)\Rightarrow (4-d)(4+d)=7\Rightarrow 4^2-d^2=7(\because x^2-y^2=(x+y\left)\right(x-y\left)\right)\Rightarrow 16-d^2=7\Rightarrow d^2=16-7\Rightarrow d^2=9\Rightarrow d=\pm \sqrt{9}\Rightarrow d=\pm 3$

Hence, the common difference is $d=\pm 3$.