How can you simplify #sqrt(28-5sqrt(12))# ?

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Answer 1 · 2016-12-12T00:14:49

#sqrt(28-5sqrt(12)) = 5 - sqrt(3)#

For a start:

#sqrt(12) = sqrt(2^2*3) = 2sqrt(3)#

So:

#sqrt(28-5sqrt(12)) = sqrt(28-10sqrt(3))#

Can this be simplified further?

Let us attempt to find rational #a, b# such that:

#28 - 10sqrt(3) = (a+bsqrt(3))^2#

#color(white)(28 - 10sqrt(3)) = (a^2+3b^2)+2a b sqrt(3)#

Equating coefficients:

#{ (a^2+3b^2 = 28), (2ab = -10) :}#

From the second equation, we find:

#b = -5/a#

Substituting #-5/a# for #b# in the first equation, we get:

#28 = a^2+75/a^2#

Subtracting #28# from both sides and multiplying by #a^2# we get:

#0 = (a^2)^2-28(a^2)+75#

#color(white)(0) = (a^2-25)(a^2-3) = (a-5)(a+5)(a-sqrt(3))(a+sqrt(3))#

Since we want #a# to be rational, we have:

#a = +-5#

If #a = 5# then #b = -5/a = -1#, resulting in #5 - sqrt(3)# which is positive.

If #a = -5# then #b = -5/a = 1#, resulting in #-5 + sqrt(3)# which is negative.

Since we want the non-negative square root, we want #5 - sqrt(3)#