Find the two values of m
There are two possible values of m such that the four roots of the equation are in arithmetic progression. If the absolute difference between these two values is p/q where p and q are coprime positive integers, what is p+q?
The answer is 631.
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1 solution
Recognise that this quartic expression can be factorised into (x^2-a^2)(x^2-b^2) where the roots are +/-a and +/-b. Assume that a<b. Then the common difference is a-(-a) = 2a. Hence b = a+2a = 3a.Now, we have (x^2-a^2)(x^2-(3a)^2) = 0. Comparing coefficients of x^2, we have a^2 + (3a)^2 = 2-m such that (1+9)a^2 = 2-m. Thus a^2= (2-m)/10. Comparing constants, we have (a^2)(3a^2) = 4m^2 so that (a * 3a)^2 = (2m)^2. Thus 3a^2 = +/- 2m. Now (3/10)(2-m) = 2m or -2m Solving these equations gives us m = 6/23 and -6/17 respectively. The absolute difference is 6/23 = 6/17 = 240/391. Therefore p+q = 240 + 391 = 631.