If # x = 3 +2sqrt(2)# Then what is #x ^ (1/2) - x ^ (-1/2)#?

1 Answer
George C.
Jul 23, 2017

#x^(1/2)-x^(-1/2) = 2#

Explanation:

Note that:

#(a+bsqrt(2))^2 = (a^2+2b^2)+2ab sqrt(2)#

Comparing the right hand expression with #3+2sqrt(2)#, note that it will match if we put #a=b=1#, so:

#(1+sqrt(2))^2 = 3+2sqrt(2)#

So #1+sqrt(2)# is a square root of #3+2sqrt(2)#. Since it is positive, it is the principal square root, so:

#(3+2sqrt(2))^(1/2) = 1+sqrt(2)#

So with #x=3+2sqrt(2)# we find:

#x^(1/2)-x^(-1/2) = (1+sqrt(2))-1/(1+sqrt(2))#

#color(white)(x^(1/2)-x^(-1/2)) = (1+sqrt(2))-(1-sqrt(2))/((1+sqrt(2))(1-sqrt(2)))#

#color(white)(x^(1/2)-x^(-1/2)) = (1+sqrt(2))-(1-sqrt(2))/(1-2)#

#color(white)(x^(1/2)-x^(-1/2)) = (1+color(red)(cancel(color(black)(sqrt(2)))))+(1-color(red)(cancel(color(black)(sqrt(2)))))#

#color(white)(x^(1/2)-x^(-1/2)) = 2#

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