Can you write #(a+b)^0.5 - (a-b)^0.5# as the square root of a difference?

2 Answers
George C.
Mar 4, 2018

#(a+b)^0.5-(a-b)^0.5 = sqrt(2a-2sqrt(a^2-b^2))#

Explanation:

If #(a+b)^0.5 - (a-b)^0.5# a.k.a. #sqrt(a+b)-sqrt(a-b)# can be written as the square root of a difference, then squaring it should give us a difference...

#(sqrt(a+b)-sqrt(a-b))^2 = (sqrt(a+b))^2-2sqrt(a+b)sqrt(a-b)+(sqrt(a-b))^2#

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

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

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

So yes, #(a+b)^0.5 - (a-b)^0.5# can be written as the square root of the difference of #2a# and #2sqrt(a^2-b^2)#, namely:

#(a+b)^0.5-(a-b)^0.5 = sqrt(2a-2sqrt(a^2-b^2))#

Cesareo R.
Mar 4, 2018

# sqrt(2a-2sqrt(a^2-b^2))#

Explanation:

#sqrt(x-y) = sqrt(a+b)-sqrt(a-b)#

squaring both sides

#x-y = 2a-2sqrt(a^2-b^2)# and then

#sqrt(x-y)=sqrt(a+b)-sqrt(a-b) = sqrt(2a-2sqrt(a^2-b^2))#

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