What is the square root of #sqrt((y^2 - z^2)(z^2 - x^2)) + sqrt((z^2 - x^2)(x^2 - y^2)) + sqrt((x^2 - y^2)(y^2 - z^2))#?

1 Answer
George C.
May 8, 2017

#sqrt(2)/2(sqrt(x^2-y^2)+sqrt(y^2-z^2)+sqrt(z^2-x^2))#

provided at least two of the following hold:

#x^2 >= y^2" "y^2 >= z^2" "z^2 >= x^2#

Explanation:

Note that:

#(x^2-y^2)+(y^2-z^2)+(z^2-x^2)#

#= color(red)(cancel(color(black)(x^2)))-color(red)(cancel(color(black)(x^2)))+color(purple)(cancel(color(black)(y^2)))-color(purple)(cancel(color(black)(y^2)))+color(violet)(cancel(color(black)(z^2)))-color(violet)(cancel(color(black)(z^2))) = 0#

So let us see what happens when we square:

#sqrt(x^2-y^2)+sqrt(y^2-z^2)+sqrt(z^2-x^2)#

as the squared terms will cancel...

#(sqrt(x^2-y^2)+sqrt(y^2-z^2)+sqrt(z^2-x^2))^2#

#=(sqrt(x^2-y^2))^2+(sqrt(y^2-z^2))^2+(sqrt(z^2-x^2))^2+2sqrt((y^2-z^2)(z^2-x^2))+2sqrt((z^2-x^2)(x^2-y^2))+2sqrt((x^2-y^2)(y^2-z^2))#

#=color(red)(cancel(color(black)((x^2-y^2)+(y^2-z^2)+(z^2-x^2))))+2sqrt((y^2-z^2)(z^2-x^2))+2sqrt((z^2-x^2)(x^2-y^2))+2sqrt((x^2-y^2)(y^2-z^2))#

#=2(sqrt((y^2-z^2)(z^2-x^2))+sqrt((z^2-x^2)(x^2-y^2))+sqrt((x^2-y^2)(y^2-z^2)))#

So the square root we want is:

#sqrt(2)/2(sqrt(x^2-y^2)+sqrt(y^2-z^2)+sqrt(z^2-x^2))#

#color(white)()#
Notes

The above answer more or less assumes that:

#sqrt(a)sqrt(b) = sqrt(ab)#

While this does hold if at least one of #a, b# is non-negative, it fails if both are negative.

This can happen in the above derivation if, for example:

#0 < x^2 < y^2 < z^2#

Then we find:

#sqrt(x^2-y^2)sqrt(y^2-z^2) = -sqrt((x^2-y^2)(y^2-z^2))#

...the opposite sign from what we need.

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