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Puzzles

10 December

The equation \(x^2+1512x+414720=0\) has two integer solutions.
Today's number is the number of (positive or negative) integers \(b\) such that \(x^2+bx+414720=0\) has two integer solutions.

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Powerful quadratics

Source: nrich
Find all real solutions to
$$(x^2-7x+11)^{(x^2-11x+30)}=1.$$

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