Maths Olympiad Prep

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Algebra Difficulty 5.0 AIME, harder Find the answer

2. Let y=(17−x)(19−x)(19+x)(17+x)y=(17-x)(19-x)(19+x)(17+x), where xx is a real number. Find the smallest possible value of yy.

Pick one

Solution

2. Answer: (A)
We have
y=(172−x2)(192−x2)=x4−(172+192)x2+172⋅192=x4−650x2+3232=(x2−325)2+3232−3252 \begin{aligned} y & =\left(17^{2}-x^{2}\right)\left(19^{2}-x^{2}\right) \\ & =x^{4}-\left(17^{2}+19^{2}\right) x^{2}+17^{2} \cdot 19^{2} \\ & =x^{4}-650 x^{2}+323^{2} \\ & =\left(x^{2}-325\right)^{2}+323^{2}-325^{2} \end{aligned}

Hence the smallest possible value of yy is 3232−3252=(−2)(648)=−1296323^{2}-325^{2}=(-2)(648)=-1296.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.