How do you show that #sqrt(2-sqrt3)/2=(sqrt6-sqrt2)/4#?

2 Answers
mason m
Jun 8, 2017

First, note that any number in the form #a-sqrtb# when squared gives a number in the form #c-sqrtd#.

In fact,

#(a-sqrtb)^2=(a^2+b)-2asqrtb=(a^2+b)-sqrt(4a^2b)#

The goal here is to reduce #sqrt(2-sqrt3)# by writing #2-sqrt3# in the form #(a-sqrtb)^2#, so the square root and the exponent of #2# will cancel one another out.

We want:

#2-sqrt3=(a-sqrtb)^2=(a^2+b)-sqrt(4a^2b)#

Creating a system:

#{(2=a^2+b),(4a^2b=3):}#

Yielding:

#{(a^2=2-b),(a^2=3/(4b)):}#

So:

#3/(4b)=2-b#

#4b^2-8b+3=0#

#(2b-1)(2b-3)=0#

#b=1/2,3/2#

And from #a^2=2-b#, these give:

#{((a,b)=(sqrt(3/2),1/2)),((a,b)=(sqrt(1/2),3/2)):}#

Which gives the identity:

#2-sqrt3=(sqrt(3/2)-sqrt(1/2))^2#

(You can verify this if you want)

And from here, we see that:

#sqrt(2-sqrt3)/2=sqrt((sqrt(3/2)-sqrt(1/2))^2)/2=(sqrt(3/2)-sqrt(1/2))/2#

#=((sqrt3-1)/sqrt2)/2=(sqrt3-1)/(2sqrt2)#

Multiplying by #sqrt2/sqrt2#:

#=(sqrt6-sqrt2)/4#

Douglas K.
Jun 8, 2017

Given: #sqrt(2-sqrt3)/2=(sqrt6-sqrt2)/4#

Multiply the left side by 1 in the form of #2/2#:

#2/2sqrt(2-sqrt3)/2=(sqrt6-sqrt2)/4#

The 2 in the numerator goes inside the square root as #2^2#

#sqrt(2^2(2-sqrt3))/4=(sqrt6-sqrt2)/4#

Distribute #4#:

#sqrt((8-4sqrt3))/4=(sqrt6-sqrt2)/4#

The 4 goes inside the inner root as 16:

#sqrt((8-sqrt48))/4=(sqrt6-sqrt2)/4#

Break 8 into 6 + 2 and factor the root term:

#sqrt((6-2sqrt6sqrt(2) + 2))/4=(sqrt6-sqrt2)/4#

Write the end terms as squares of roots:

#sqrt(((sqrt6)^2-2sqrt6sqrt(2) + (sqrt2)^2))/4=(sqrt6-sqrt2)/4#

It fits the pattern of a square of the difference:

#sqrt((sqrt6-sqrt(2))^2)/4=(sqrt6-sqrt2)/4#

Square root of a square

#(sqrt6-sqrt(2))/4=(sqrt6-sqrt2)/4#

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