8
$\begingroup$

I came across this olympiad algebra problem, asking to solve for $x$:

$x\ +\ \sqrt { (x+1)(x+2) } \ +\ \sqrt { (x+2)(x+3) } +\ \sqrt { (x+3)(x+1) } =\ 4$

Here was my try:

If $$x\ +\ \sqrt { (x+1)(x+2) } \ +\ \sqrt { (x+2)(x+3) } +\ \sqrt { (x+3)(x+1) } =\ 4$$

Then $\quad \sqrt { (x+1)(x+2) } +\sqrt { (x+2)(x+3) } +\sqrt { (x+3)(x+1) } =4-x$.

Further, I tried squaring the equation on both sides, but that doesn't seem to solve my problem. Please help.

Thank you.

$\endgroup$
  • 1
    $\begingroup$ The octic method described in my answer can be used to show that for any rational constant $a,b,c$, then $$(x+a)(x+b) = y_1^2\\(x+a)(x+c) = y_2^2\\(x+b)(x+c) = y_3^2$$ has an infinite number of rational solutions $x$. $\endgroup$ – Tito Piezas III Dec 20 '15 at 17:55
6
$\begingroup$

While Kf-Sansoo has given an elegant answer, if the problem asks for any real (and not just rational) solution, then it misses a second one which is a root of a quartic equation, hence normally is not easy to do by hand.

In general, the two solutions to,

$$x+\sqrt{(x+1)(x+2)}+\sqrt{(x+2)(x+3)}+\sqrt{(x+3)(x+1)} = n\tag1$$

for real $n>0$ are,

$$x = \frac{(n^2+4n+5)^2}{4(n+1)(n+2)(n+3)}-2\tag{2a}$$

and the appropriate root of,

$$-23 + 48 n - 22 n^2 + n^4 - 4 (30 - 33 n + 6 n^2 + n^3) x \\+ 16 (-11 + 6 n) x^2 + 16 (-6 + n) x^3 - 16 x^4=0\tag{2b}$$

For $n = 4$, we have $x_1 = -311/840 \approx -0.37$. Then $x_2 \approx -5.12357$ as a root of,

$$73 - 232 x + 208 x^2 - 32 x^3 - 16 x^4 = 0$$

with both valid for the positive case of $\sqrt{z}$ as the graph from Walpha below shows,

enter image description here


$\color{green}{Edit:}$ When using Kf-Sansoo's method, we end up with an expression of form,

$$\prod^4 (c_1\sqrt{x+1}\pm c_2\sqrt{x+2}\pm c_3\sqrt{x+3}) = 0$$

Let $n=4$ and we get $x = -311/840$. Simpler, but the price to pay is we lose a second solution. Another method is to form an octic,

$$\prod^8 \big(y-(\pm\sqrt{z_1}\pm \sqrt{z_2}\pm \sqrt{z_3})\big)=0 \tag3$$

After it is formed, substitute into $(3)$ the ff,

$$y = n-x\\z_1=(x+1)(x+2)\\z_2=(x+2)(x+3)\\z_3=(x+3)(x+1)$$

and we get linear/quartic factors given by $(2a), (2b)$. Less simpler, but it yields a second valid solution.

$\endgroup$
  • $\begingroup$ Just by curiosity, how did you arrive to thse general expressions ? Are they valid for any $n$ (real or integer only ?). Thanks $\endgroup$ – Claude Leibovici Dec 20 '15 at 6:33
  • $\begingroup$ @ClaudeLeibovici: I used the octic formed by $y-(\pm\sqrt{z_1}\pm\sqrt{z_2}\pm\sqrt{z_3})$, hence to completely solve the problem by hand is too much. The two solutions should be valid for any real $n$, provided one uses the correct signs of $(1)$ and the appropriate root of $(2b)$. $\endgroup$ – Tito Piezas III Dec 20 '15 at 6:41
  • $\begingroup$ Thank you ! It is impressive. I think that it could be good you add these explanations in your answer. Cheers. $\endgroup$ – Claude Leibovici Dec 20 '15 at 6:47
  • $\begingroup$ @ClaudeLeibovici: And I've added a graph which shows the two solutions. $\endgroup$ – Tito Piezas III Dec 20 '15 at 7:27
  • $\begingroup$ Thank you ! You improved a lot !! $\endgroup$ – Claude Leibovici Dec 20 '15 at 7:30
9
$\begingroup$

Hint: Observe that $:x = (x+1) + (x+2) - (x+3)$. This leads us to letting $a = \sqrt{x+1}, b = \sqrt{x+2}, c = \sqrt{x+3} \Rightarrow a^2+b^2-c^2 + ab+bc+ca = 4, b^2-a^2 = 1 = c^2-b^2 \Rightarrow a^2-1 +ab+bc+ca= 4 \Rightarrow (a+b)(a+c) = 5 \Rightarrow \dfrac{a+c}{b-a} = 5 \Rightarrow a+c = 5b-5a\Rightarrow 6a = 5b-c\Rightarrow 6\sqrt{x+1} = 5\sqrt{x+2} - \sqrt{x+3}$. Can you continue?

$\endgroup$
  • 1
    $\begingroup$ There is a second solution. Kindly see graph in my answer. $\endgroup$ – Tito Piezas III Dec 20 '15 at 7:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.