Question

If the roots of $x^3-9x^2+23x-15=0$ are in A.P, then the common difference of A.P. is

• A. $\pm 5$
• B. $\pm 4$
• C. $\pm 3$
• D. $\pm 2$

Answer 1

Answer: (D)

$\pm 2$

Let the roots be $a-d,a,a+d$ where d is the common difference.

Hence the sum of roots is

$(a-d)+a+(a+d)=\frac{-B}{A}$

$3a=\frac{-(-9)}{1}$

$3a=9$

$a=3$ ...(i)

And product of roots is

$(a-d)(a+d)\left(a\right)=\frac{-D}{A}$

$(a^2-d^2)a=\frac{-(-15)}{1}$

$a^3-a{d}^2=15$

Now $a=3$.

Hence

$27-3d^2=15$

$9-d^2=5$

$4=d^2$

$d=\pm 2$.

Hence the common difference is $\pm 2$.