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.
Best Answer
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.