<s>[INST] How do I solve $(x-1)(x-2)(x-3)(x-4)=3$ How to solve $$(x-1) \cdot (x-2) \cdot (x-3) \cdot (x-4) = 3$$
Any hints?
 [/INST] I assume the hints would already have given you the answer. If not, here is the full answer:
Let y = x-2.5
(y+1.5)(y-1.5)(y+0.5)(y-0.5) = 3.
so $(y^2-2.25)(y^2-0.25) = 3$.
Let $z = y^2-1.25$.
(z-1)(z+1) = 3.
So $z^2-1 = 3$.
Hence $z^2 = 4$.
z = -2 gives $y^2 = z + 1.25 = -0.75$.
So $y = \pm \sqrt{0.75}i$. Clearly, this should be ignored if you only want real roots.
$x = 2.5 + y = 2.5 \pm \sqrt{0.75}i$.
z = 2 gives $y^2 = z + 1.25 = 3.25$.
So $y = \pm \sqrt{3.25}$. Clearly, this should be ignored if you only want real roots.
$x = 2.5 + y = 2.5 \pm \sqrt{3.25}$.
If you want all the terms in the product to be positive, then obvious ly $x = 2.5 + \sqrt{3.25}$ is the only one that works. This is roughly 4.30277564.
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