Question 387914: The roots of the equation x^2+px+1=0 are a and b and the roots of the equation x^2-9x+q=0 are a+2b and 2a+b. Determine the value of p and q.
Answer by robertb(5830)   (Show Source):
You can put this solution on YOUR website! From the given, it must be that ab = 1 (from the 1st given) and -(-9) = a+2b + 2a+b, or 9 = 3a + 3b, or a+b = 3 (from the 2nd given).
Thus we have the system
ab = 1
a+b = 3.
Hence we must have -p = a+b = 3, or p = -3. (From the middle coefficient of x^2+px+1=0).
Now 2a + 2b = 6, and so
a+2b = 6-a, and 2a + b = 6 - b.
Multiplying these two quantities, we must get (a+2b)(2a+b) = 36 - 6a - 6b + ab = 36-6(a+b) + ab = 36 - 6*3 + 1 = 19. Now by theorem (a+2b)(2a+b) = q, and therefore q = 19.
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