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

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Answer 1 · 2018-02-12T22:24:15

#47#

Given:

#sqrt(x)+1/sqrt(x) = 3#

Square both sides to get:

#x+2+1/x = 9#

Subtract #2# from both sides to get:

#x+1/x = 7#

Square both sides to get:

#x^2+2+1/x^2 = 49#

Subtract #2# from both sides to get:

#x^2+1/x^2 = 47#

Details

Note that:

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

So we find:

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

Similarly:

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