Question

If the roots of $x^4+q{x}^2+kx+225=0$ are in arithmetic progression, then the value of q is

• A. $15$
• B. $25$
• C. $35$
• D. $50$

Answer 1

The correct option is D $50$

Given that the roots are in A.P

Let $a-3d,a-d,a+d,a+3d$ be the four roots of the quartic equation

Sum of the roots$=a-3d+a-d+a+d+a+3d=4a=0$

$\Rightarrow a=0$

Sum of the roots taken two at a time

$=(a-3d)(a-d)+(a-d)(a+d)+(a+d)(a+3d)+(a+3d)(a-3d)+(a-3d)(a+d)+(a-d)(a+3d)=q$

$\Rightarrow 3d^2-d^2+3d^2-9d^2-3d^2-3d^2=q$ since $a=0$

$\Rightarrow -10d^2=q$

Product of the roots$=(a-3d)(a-d)(a+d)(a+3d)=125$

$\Rightarrow 9d^4=125$ since $a=0$

$\Rightarrow d^4=\frac{125}{9}=25$

$\therefore d^2=\pm 5$

Again $-10d^2=q$

$\Rightarrow -10\times \pm 5=q$

$\therefore q=-50$ or $50$