Determine the real values of q so that the equation given by x
4-40*x
2+q=0 has real solutions forming an arithmetic sequence.
The equation can be factored into (x^2-a^2)*(x^2-b^2) for some constants a and b This means a^2+b^2=40 and a^2*b^2=q. Also the roots will be a, -a, b, and -b. Without loss of generality assume a is closer to 0 than b.
Then the roots in order are -b, -a, a, b. This is to be an arithmetic sequence. The common difference is given by a-(-a)=2a. Then b-a=2a, which makes b=3a.
Substitute this into a^2+b^2=40. Then 10a^2=40, which makes a=2 or a=-2. In either case a^2=4 and then b^2=36. Then q=4*36=144.
Solution
