\[ \sqrt{a-b} + \sqrt{b-c} + \sqrt{c-d} + \sqrt{d-a} = K \] for some real constant $K$ and some real numbers $a$, $b$, $c$ and $d$. Find $K$.

Again i apologize for the syntax and appreciate anyhelp thanks!

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    $\begingroup$ Hint: For the first $\sqrt{}$ to be defined, you need $a \ge b$ ... for the second ... $\endgroup$ – martini Jul 12 '12 at 7:18
  • $\begingroup$ @Daniel: As martini noted nicely, what will happen if $a≥b≥c≥d≥a$? $\endgroup$ – mrs Jul 12 '12 at 7:29

Here $a, b,c,d,K$ are real numbers.

$\sqrt{a-b} , \sqrt{b-c} , \sqrt{c-d} , \sqrt{d-a}$ are real numbers because K is real number.

So,$\sqrt{a-b}$ real numbers means $ a \geq b$ and rest of also same as thing as.

That is, $ a \geq b, b \geq c, c \geq d, d \geq a$

This implies $a=b=c=d$, The value of K should be zero.

  • $\begingroup$ Is it really immediately clear that all four roots are real? A priori some of them might be imaginary such that the sum is real. $\endgroup$ – Simon Markett Jul 12 '12 at 10:14
  • $\begingroup$ Here a,b,c,d are real numbers and hence K is zero. In this case all roots are zeros. $\endgroup$ – Prasad G Jul 12 '12 at 10:22
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    $\begingroup$ Huh, yh I guess the square root isn't really well defined on negative numbers anyway... makes sense. $\endgroup$ – Simon Markett Jul 12 '12 at 10:31

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