# 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?

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Look at $(x-1)(x-4)$ and $(x-2)(x-3)$ they multiply as $(x^{2}-5x+4)$ and $x^{2}-5x+6$. Now put $t= x^{2}-5x$ and reduce it to a quadratic equation.

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Alternately, $z = x-\frac{5}{2}$, then the equation becomes $(z^2-\frac{1}{4})(z^2-\frac{9}{4})=3$. But this is basically the same solution.... –  N. S. Apr 16 '12 at 16:31
Note that solving said equation for $t$ yields roots $t = -3, -7.\:$ Then you solve $\:x^2-5x = t,\:$ which amounts to solving the two quadratics I gave. Rather than this roundabout way, it's simpler to notice that subtracting $3$ preserves the difference of squares form, as in my answer. –  Bill Dubuque Apr 16 '12 at 17:13
@N.S.: Equivalent it may be, but your substitution $z=x-\frac{5}{2}$ is a natural move, in that it brings out the symmetry. –  André Nicolas Apr 16 '12 at 17:30
Several solution methods (and some motivation behind them) for the equation $(x-r)(x-2r)(x-3r)(x-4r) = a$ are given in the following sci.math thread: groups.google.com/group/sci.math/browse_thread/thread/… (Google) mathforum.org/kb/message.jspa?messageID=6344054 (Math Forum) –  Dave L. Renfro Apr 16 '12 at 18:11
To elaborate on what N.S. and André are saying: that substitution depresses the quartic, in that it kills the cubic term, and very luckily, the linear term as well... –  Ｊ. Ｍ. Apr 17 '12 at 11:31

Hint $\$ The LHS is a difference of squares $\rm\:y^2\!-\!1,\:$ hence so too is $\rm\:(y^2\!-\!1)-3\: =\: y^2\!-\!2^2,\:$ viz.

$\rm\qquad\ \:\! (x\!-\!1)(x\!-\!4) (x\!-\!2)(x\!-\!3)\ =\ (x^2\!-\!5x+4)(x^2\!-\!5x+6)\ =\ (x^2\!-\!5x+5)^2 \!-\! 1^2$

$\rm\ \ \Rightarrow\ (x\!-\!1)(x\!-\!4) (x\!-\!2)(x\!-\!3)\!-\!3\ =\ (x^2\!-\!5x+5)^2 \!-\! 2^2\ =\ (x^2\!-\!5x+3)(x^2\!-\!5x+7)$

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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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In writing mathematics, once should be encouraged to use rationals 3/2 and not decimals 1.5 ... but of course rationals are less convenient for on-line writing. –  GEdgar Apr 17 '12 at 12:19