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.
Start by finding the positive real solutions to:
y^2 - 40y + q = 0
x^2 = y = 20 ± sqrt(400 - q)
So q <= 400 and if we need 4 real x solutions, q >= 0
There will be two positive roots and two mirror image negative roots.
For all 4 x solutions to be in arithmetic sequence, the larger positive root must be 3 times the smaller positive root, so the pattern of 4 roots will be (-3k, -k, +k, +3k) to put an equal gap between each two adjacent roots.
For the moment, let sqrt(400 - q) = u
We want sqrt(20+u) = 3*sqrt(20-u)
(20+u) = 9(20-u)
10u = 160
u = 16
sqrt(400 - q) = 16
(400 - q) = 256
q = 144
x^4 - 40x^2 + 144 = 0
which has 4 roots in arithmetic sequence
(-6,-2,2,6)
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Posted by Larry
on 2024-07-02 11:46:58 |
solution
