Find the positive solution to 1x2−10x−29+1x2−10x−45−2x2−10x−69=0
Let n be the smallest positive integer that is a multiple of 75 and has exactly 75 positive integral divisors, including 1and itself. Find n/75
1.
Let a=x2−10x−29
1a+1a−16−2a−40=0.(a−16)(a−40)+a(a−40)−2(a)(a−16)=0−64a+40×16=0,⇒a=10.
10=x2−10x−29⟺0=(x−13)(x+3) The positive root is 13.
2.
https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_5
I hope this helped,
Gavin
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