2020/017) FInd q when the equation $x^4-40x^2+q=0$ has 4 roots is AP.

Because it has 4 roots in AP so let the roots be $a-3d, a-d,a+d, a+3d$ 
The sum of the root is zero as coefficient of $x^3=0$
So we have $(a-3d)+(a-d)+(a+d)+(a-3d) = 4a = 0$
Or $a=0$
Hence the roots are $-3d, -d, d, 3d$
So Equation becomes  $(x+3d)(x+d)(x-d)(x-3d)=0$
Or $(x+3d)(x-3d)(x+d)(x-d)=0$
Or $(x^2-9d^2)(x^2-d^2) =  x^4-10d^2x^2+ 9d^4=0$
Comparing with given equation $-10d^2= - 40$ or $d^2=4$
And $q=9d^4= 9 (4)^2= 144$
Hence $q= 144$