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How many solutions the following equation has in real numbers ?$$\sqrt{\frac{x-8}{1388}}+\sqrt{\frac{x-7}{1389}}+\sqrt{\frac{x-6}{1390}}=\sqrt{\frac{x-1388}{8}}+\sqrt{\frac{x-1389}{7}}+\sqrt{\frac{x-1390}{6}}$$

$1)1\quad\quad\quad\quad\quad\quad\quad2)2\quad\quad\quad\quad\quad\quad\quad3 )3\quad\quad\quad\quad\quad\quad\quad4)4\quad\quad\quad\quad\quad5)\text{zero}$

Obviously we have $x\ge1390$. starting from $x=1390$ we can see LHS is very close to $1+1+1=3$ and RHS is close to zero. for higher values of $x$ we can see right side of the equation grows much faster than left side. so it seems the equation has not answer in real numbers.

I'm not sure how to proceed mathematically. I can rewrite it as follow: $$\left(\sqrt{\frac{x-8}{1388}}-\sqrt{\frac{x-1388}{8}}\right)+\left(\sqrt{\frac{x-7}{1389}}-\sqrt{\frac{x-1389}{7}}\right)+\left(\sqrt{\frac{x-6}{1390}}-\sqrt{\frac{x-1390}{6}}\right)=0$$ I putted fractions with similar numbers in the same parentheses so it looks better now, but don't know how to continue.

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    $\begingroup$ If the RHS 'starts' (at $x=1390$) smaller than the left and grows more quickly, then that's surely a sign that the two cross paths at some point. Can you see how you might be able to put 'grows more quickly' into formal language and then use that to answer the question? $\endgroup$ Mar 12 '21 at 20:32
  • $\begingroup$ @StevenStadnicki should we find derivative of both side of the equation and comparing them? $\endgroup$
    – Amirali
    Mar 12 '21 at 20:36
  • $\begingroup$ Perhaps it helps to notice that $x=1396$ makes each square root reduce to $1$? Though maybe that's not enough. Certainly it eliminates "zero" from the given choices. $\endgroup$
    – user170231
    Mar 12 '21 at 21:05
  • $\begingroup$ Given the algebra-precalculus tag, I would surmise that derivatives are outlawed. If this question is intended for math students who have not taken Calculus, then the problem composer must have intended that the solution would not require Calculus. $\endgroup$ Mar 12 '21 at 21:35
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Write $x=1396+t$. Then $$\sqrt{1+\frac t{1388}}\:+\sqrt{1+\frac t{1389}}\:+\sqrt{1+\frac t{1390}}$$ $$=\sqrt{1+\frac t8}\:+\sqrt{1+\frac t7}\:+\sqrt{1+\frac t6}.$$ Clearly $t=0$ is one solution. Now, $t$ cannot be positive, because that would make the terms on the LHS each smaller than the terms on the RHS. Similarly, $t$ cannot be negative, for the opposite reason. Therefore $t=0$ or $x=1396$ is the unique solution.

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  • $\begingroup$ My sincere compliments for an answer that I have upvoted. $\endgroup$
    – Sebastiano
    Mar 13 '21 at 13:13
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$$\sqrt{\frac{x-8}{1388}}+\sqrt{\frac{x-7}{1389}}+\sqrt{\frac{x-6}{1390}}=\sqrt{\frac{x-1388}{8}}+\sqrt{\frac{x-1389}{7}}+\sqrt{\frac{x-1390}{6}} \tag {1}$$

Since $x \geq 1390$, let us put $x = 1390 + k$ where $k \geq 0$

$(1) \implies$

$$\sqrt{\frac{1382+k}{1388}}+\sqrt{\frac{1383+k}{1389}}+\sqrt{\frac{1384+k}{1390}}=\sqrt{\frac{k+2}{8}}+\sqrt{\frac{k+1}{7}}+\sqrt{\frac{k}{6}}$$

Note that since $k \geq 0$,

we have $$\left(\sqrt{\frac{1382+k}{1388}} < \sqrt{\frac{1383+k}{1389}} < \sqrt{\frac{1384+k}{1390}} < 1 \right) \land \left(\sqrt{\frac{k}{6}} < \sqrt{\frac{k+1}{7}} < \sqrt{\frac{k+2}{8}} < 1 \right)$$ for $k < 6$,

$$\left(\sqrt{\frac{1382+k}{1388}} > \sqrt{\frac{1383+k}{1389}} > \sqrt{\frac{1384+k}{1390}} > 1 \right) \land \left(\sqrt{\frac{k}{6}} > \sqrt{\frac{k+1}{7}} > \sqrt{\frac{k+2}{8}} > 1 \right)$$ for $k > 6$ and

$$\left(\sqrt{\frac{1382+k}{1388}} = \sqrt{\frac{1383+k}{1389}} = \sqrt{\frac{1384+k}{1390}} = 1 \right) \land \left(\sqrt{\frac{k}{6}} = \sqrt{\frac{k+1}{7}} = \sqrt{\frac{k+2}{8}} = 1 \right)$$ for $k = 6$

Can you argue why the first two cases do not lead to any solution but the last case does?

In fact, you can show that

$\left(\sqrt{\frac{k}{6}} < \sqrt{\frac{k+1382}{1388}} \right) \land \left(\sqrt{\frac{k+1}{7}} < \sqrt{\frac{k+1383}{1389}} \right) \land \left(\sqrt{\frac{k+2}{8}} < \sqrt{\frac{k+1384}{1390}} \right)$ for $k < 6$

and

$\left(\sqrt{\frac{k}{6}} > \sqrt{\frac{k+1382}{1388}} \right) \land \left(\sqrt{\frac{k+1}{7}} > \sqrt{\frac{k+1383}{1389}} \right) \land \left(\sqrt{\frac{k+2}{8}} > \sqrt{\frac{k+1384}{1390}} \right)$ for $k > 6$

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The easiest but absurd way to do this is by squaring and squaring and squaring and..... To write out the equation of polynomial in $x$

$$\sqrt{\frac{x-8}{1388}}+\sqrt{\frac{x-7}{1389}}+\sqrt{\frac{x-6}{1390}}-\sqrt{\frac{x-1388}{8}}-\sqrt{\frac{x-1389}{7}}-\sqrt{\frac{x-1390}{6}} = 0$$

With the help of a cas, then we would have to observe the polynomial in $x$ to see how many real roots it has

$$( 166271366055421398492452275393603941230766862954455137380647248140804636783475714327462432993480052899307496588105274122048146156780410637021361 \cdot x^6 - 1420140036782905319437561361639895711607700448706531408943829995316272533767501640030686352288245638342331131237991828557619781758027571151457168696 \cdot x^5 + 5053362485747062046204356842284861702806221919004319273920353418806545439134535332974351841718762407835957590697696948539773086891685978542298184835440 \cdot x^4 - 9589047098047637215433435077525811816830831747049585209634085291233160531678815568240751343497601637440011806189213827409633807381969946623370920034699520 \cdot x^3 + 10233876003972506215993189306355354430848302521410622385212634701164151435860483711557613487835971006731199476091003849691449303854798610976036031554449411840 \cdot x^2 - 5824396568911083719726988788037339915692680410889570986360670278090638276801032058096056401851613752306491371113890184533047015490065483972816831803846901282816 \cdot x + 1381013855402990033245846225695823841737429380881009697772601168066108621988351698959635167325016829640478323081178371168331795718011254016938645598239579277398016 )\cdot (x-1396)= 0 $$

The polynomial is $7$th degree, and has $4$ real roots

EDIT:

But squaring an equation creates an environment for false root.... so all roots are tested, so that only $x-1396 = 0$ is a real root that satisfies the equation

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