Question

if a^2+b^2+1/a^2+1/b^2=4, then the value of √(a^2 +b^2)

Answer 1

Step-by-step explanation:

The answer is 0(Zero).

a^2+b^2+a^-2+b^-2 = 4

=> ((a^2+a^(-2) – 2)+(b^2+b^(-2) -2))+4 = 4

=> (a-(1/a))^2 + (b-(1/b))^2 = 0

=> a - (1/a) = 0 and b - (1/b) = 0

=> a = 1/a and b = 1/b

=> a^2 = 1 and b^2 = 1

So a^2 - b^2 = 0.

I hope it's helpful.

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