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2001 SMT/Algebra Problems/Problem 9

Problem

Find all solutions to $(x - 3)(x - 1)(x + 3)(x + 5) = 13$.

Solution

Substitute $y=x+1$. Then the equation becomes \[(y-4)(y-2)(y+2)(y+4)=13\] Combine using difference of squares, and let $z=y^2$: \[(z-16)(z-4)=13\] \[z^2-20z+51=0\] Thus $z=3$ or $z=17$. Going back, $y=\pm\sqrt{3},\pm\sqrt{17}$, so the solution set of $x$ is: \[x\in\{-\sqrt{3}-1,\sqrt{3}-1,-\sqrt{17}-1,\sqrt{17}-1\}\]

~ eevee9406

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