Question

how to solve.....................​

Answer 1

if you follow the putting value system ...

RHS =4

so LHS should be 4

so x²+y² +1/x² +y² must be =4

so each term in this case =1

x²=1, y²=1 etc ....

so

x²+y² =2

hope this help you ...

you can put the value and check you answer ... no need to solve in long process ..

thanks !!

Answer 2

Answer:

2

Step-by-step explanation:

Given :

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

Find, x² + y²²

Solution :

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

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

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

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

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

For the sum of two positive real numbers to be 0,

both must be 0,.

⇒ $(x-\frac{1}{x} )^{2} = 0$ &  $(y-\frac{1}{y})^{2} = 0$

⇒  $x-\frac{1}{x} = 0$ &  $y-\frac{1}{y} = 0$

⇒ $x = \frac{1}{x}$ & $y = \frac{1}{y}$

⇒ $x^2 = 1$ & $y^2 = 1$

⇒ $x^2 + y^2 = 1 + 1 = 2$