Try this beautiful problem from the PRMO, 2017 based on Roots and coefficients of equations.
Let a,b be integers such that all the roots of the equation \((x^{2}+ax+20)(x^{2}+17x+b)\)=0 are negetive integers, find the smallest possible values of a+b.
Polynomials
Roots
Coefficients
Answer: is 25.
PRMO, 2017, Question 4
Polynomials by Barbeau
\((x^{2}+ax+20)(x^{2}+17x+b)\)
where sum of roots \( \lt \) 0 and product \( \gt 0\) for each quadratic equation \(x^{2}\)+ax+20=0 and
\((x^{2}+17x+b)=0\)
\(a \gt 0\), \(b \gt 0\)
now using vieta's formula on each quadratic equation \(x^{2}\)+ax+20=0 and \((x^{2}+17x+b)=0\), to get possible roots of \(x^{2}\)+ax+20=0 from product of roots equation \(20=(1 \times 20), (2 \times 10), (4 \times 5)\)
min a=4+5=9 from all sum of roots possible
again using vieta's formula, to get possible roots of \((x^{2}\)+17x+b)=0 from sum of roots equation \(17=-(\alpha + \beta) \Rightarrow (\alpha,\beta)=(-1,-16),(-2,-15),\)
\((-8,-9)\)
min b=(-1)(-16)=16 from all products of roots possible
\((a+b)_{min}=a_{min}+b_{min}\)=9+16=25.
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