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I recently participated in a mathematics tournament and encountered a question asking for the sum of the real solutions of the equation $$(x-7)(x-3)(x+5)(x+1)=1680$$

As the math tournament was time constrained and I only knew how to solve this equation by creating a new fourth degree expression and then subtracting $1680$ from that expression then refactoring to find the real roots. However after coming home I did work this out and got the two roots of $x=-7$ and $x=9$ making the sum $2$ which aligns with the answer key.

However, I am still left wondering how this could be done in a more timely manner and my research on the topic has yielded few results.

TShiong
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Eli Compton
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  • Hint: $,-3 + 1 = -7 + 5,$. – dxiv Mar 25 '23 at 21:12
  • Another Hint: take $x-7$ as $a$ – Aadi Mar 25 '23 at 21:26
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    Really the most useful substitution is $y=x-1$ in which case the equation turns into $(y-6)(y-2)(y+6)(y+2)=1680$, i.e. $(y^2-36)(y^2-4)=1680$. Now another substitution $z=y^2$, solve the quadratic equation etc. –  Mar 25 '23 at 21:28
  • Related: If $(x+1)(x+3)(x+5)(x+7)= 5760$, what are the possible values of $x$?, Solving a polynomial in an easier manner. For the problem here $,\frac{(x-3)+(x+1)}{2} = \frac{(x-7)+(x+5)}{2} = x - 1,$ so the LHS is symmetric across $x=1$. – dxiv Mar 25 '23 at 21:33
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    In fact, the symmetry is what solves the problem. Namely, if $P(x)=P(-x)$ for a real (or complex!) polynomial $P$ , it is easy to see that $P(x)=Q(x^2)$ for a simpler polynomial $Q$. Namely, if $P(x)=a_0+a_1x+a_2x^2+\ldots+a_nx^n$, then $P(-x)= a_0-a_1x+a_2x^2-\ldots\pm a_nx^n$ and this implies $a_1=-a_1, a_3=-a_3,\ldots$ i.e. all the odd coefficients are zero and so $P(x)=Q(x^2)$ with $Q(x)=a_0+a_2x+a_4x^2+\ldots$. Worth knowing! –  Mar 25 '23 at 21:38
  • $1680$ is easily factored as $2^4\times 3\times 5\times 7$ and all four factors in the equation have the same parity, so one of them must be $\pm 2$. Do we really need to solve a polynomial for this? – Daniel Mathias Mar 25 '23 at 22:46
  • @DanielMathias That's assuming integer $x$, which is good heuristics but not a proof proper. – dxiv Mar 25 '23 at 23:54

2 Answers2

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Consistent with the comment of dxiv,
let $~y = x-1,~$ and let $~u = y^2.$

Then

$$(y^2 - 4)(y^2 - 36) = 1680 \implies $$

$$u^2 - 40u - 1536 = 0 \implies \tag1 $$

$$u = \frac{1}{2} ~\left[40 \pm \sqrt{1600 + (4 \times 1536)}\right] \tag2 $$

$$ = \frac{1}{2} \left[40 \pm 88\right] = 20 \pm 44.$$

Since you are looking for real solutions, you must have that

$$y^2 = u = 64 \implies (x-1) = y = \pm 8.$$

Addendum

Actually, all of the Math, following (2) above is unnecessary. From (2) above, it is immediate that the value inside the radical will be bigger than $~(40)^2,~$ and so only one (positive) value of $~y^2 = u~$ will be forthcoming.

Then regardless of the eventual computed value of $~u = y^2,~$ you will have the two real roots of

$$x-1 = y = \pm \sqrt{u}.$$

So, denoting the two roots as $~x_1,~$ and $~x_2,~$ you have that $~\displaystyle (x_1 - 1) + (x_2 - 1) = \sqrt{u} + (-\sqrt{u}) = 0.$

So, without bothering to solve for $~u,~$ you know that $~x_1 + x_2 = 2.$

In fact, with experience, you can see that because the third term in (1) above, $~-1536,~$ is negative, you could (instead) stop the Math after (1) above, rather than (2) above.

That is, (1) above, with any negative 3rd term, is sufficient to imply that only one positive real root of $~y^2 = u~$ will be forthcoming.

user2661923
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I would solve the problem in the following way: Define $t = x^2 - 2x$, then this quadratic has a real solution if $t > -1$, and in that case, sum of the real roots is $2$. Having this in mind, simplify the expression as below: $$ \left[ (x-7)(x+5)\right] \left[(x-3)(x+1) \right] = (t-35)(t-3) = 1680$$ from which it is clear that two such values of $t$ exsits, a larger positive number and a large negative number. Thus, the solution is $2$. You don't need to solve for the roots of the polynomial which saves a lot of time.

Sam
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