Question

If x^2+y^2+1/x^2+1/y^2=4 then find x^2+y^2

Answer 1

x² +y² + 1/x² +1/y² =4
⇒(x² +1/x²-2) +(y²+1/y²-2) =0
⇒(x-1/x)² +(y-1/y)² =0
we know that square terms are always positive and sum of two positive terms can be zero only if the terms are zero.
That means,
(x-1/x) =0 and (y-1/y) =0
⇒x²-1=0
⇒x = 1 or -1
and, y²-1=0
⇒y =1 or -1
∴x²+y² = 1+1 =2

Answer 2

Answer:

$\boxed{x^2+y^2 = 2}$

Step-by-step explanation:

Given:

$x^2 + y^2 + \frac{1}{x^2} +\frac{1}{y^2} = 4$

$\implies x^2 + \frac{1}{x^2} +y^2 +\frac{1}{y^2} - 4 = 0$

$\implies x^2 + \frac{1}{x^2} - 2 + y^2 +\frac{1}{y^2} - 2 =0$

$\implies ( x - \frac{1}{x})^2 + (y-\frac{1}{y})^2 = 0$

A square is always a positive number.

If the sum of two positive is zero this means that both numbers are zero.

Thus:

$x-\frac{1}{x}=0$

$\implies x=\frac{1}{x}$

$\implies x^2 = 1$.................(1)

$y-\frac{1}{y}=0$

$y=\frac{1}{y}$

$\implies y^2=1$.......................(2)

Adding (1) and (2) we get:

$x^2 + y^2 = 1+1 \implies 2$

Your answer is 2

Hope it helps you

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