Maths Olympiad Prep

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, 2020

Algebra Difficulty 5.0 AIME, harder Prove it United States

Problem:
For a real number rr, the quadratics x2+(r−1)x+6x^{2}+(r-1)x+6 and x2+(2r+1)x+22x^{2}+(2r+1)x+22 have a common real root. The sum of the possible values of rr can be expressed as ab\frac{a}{b}, where a,ba, b are relatively prime positive integers. Compute 100a+b100a+b.

Solution

Solution:
Let the common root be ss. Then,
s2+(r−1)s+6=s2+(2r+1)s+22 s^{2}+(r-1)s+6 = s^{2}+(2r+1)s+22
and s=−16r+2s = -\frac{16}{r+2}. Substituting this into s2+(r−1)s+6=0s^{2}+(r-1)s+6=0 yields
256(r+2)2−16(r−1)r+2+6=0 \frac{256}{(r+2)^{2}} - \frac{16(r-1)}{r+2} + 6 = 0
After multiplying both sides by (r+2)2(r+2)^{2}, the equation becomes
256−16(r−1)(r+2)+6(r+2)2=0 256 - 16(r-1)(r+2) + 6(r+2)^{2} = 0
which simplifies into
5r2−4r−156=0 5r^{2} - 4r - 156 = 0
Thus, by Vieta's Formulas, the sum of the possible values of rr is 45\frac{4}{5}.

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