Question

If x^2+y^2+(1÷x^2)+(1÷y^2)=4, then find the value of x^2+y^2
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Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Answer:

First we will re-write into two parts.

==> ( x^2 + 1/x^2 ) + ( x+ 1/x) = 4

Let x + 1/x = y..............(1)

==> (x+ 1/x)^2 = y^2

=> x^2 + 2 + 1/x^2 = y^2

==> x^2 + 1/x^2 = y^2 - 2............(2)

Now we will substitute ( 1) and (2).

==> (y^2 - 2 ) + ( y) = 4

==> y^2 + y - 2 = 4

==> y^2 + y - 6 = 0

==> ( y + 3) ( y-2) = 0

==> y1= -3

==> x + 1/x = -3

==> x^2 + 1 = -3x

==> x^2 + 3x + 1 = 0

==> x1 = ( -3 + sqrt(5) /2

==> x2= ( -3-sqrt5)/2

==> y2= 2

==> x+ 1/x = 2

==> x^2 + 1 = 2x

==> x^2 -2x + 1 =0

==> ( x-1)^2 = 0

==> x= 1

Then x values are: