If #sqrt(x) + 1/sqrt(x) = 3#, what is #x^2 + 1/x^2#?
1 Answer
Feb 12, 2018
Explanation:
Given:
#sqrt(x)+1/sqrt(x) = 3#
Square both sides to get:
#x+2+1/x = 9#
Subtract
#x+1/x = 7#
Square both sides to get:
#x^2+2+1/x^2 = 49#
Subtract
#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#