Question

If x^2+y^2=29and xy=2 , find the value of x+y andx-y

Answer 1

x² + y² = 29

xy = 2

[ ( x + y )² = x² + y² + 2xy ]

( x + y )² = 29 + 2 ( 2 )

( x + y )² = 29 + 4 = 33

__χ + у = √33__

Now,

[ ( x - y )² = x² + y² - 2xy ]

( x - y )² = 29 - 2 ( 2 )

( x - y )² = 29 - 4 = 25

x - y = √25

__χ - у = 5__

Answer 2

Answer:

Given: x^2 + y^2 = 29

xy = 2

1st....

x+y

Let x+y = 0

There fore (x+y)^2 = 0^2

x^2 + y^2 + 2xy=0

29 + 2×2 = 0

29 + 4 =0

=33

2nd......

x-y

Let x - y =0

Therefore (x-y)^2 = 0^2

x^2 + y^2 -2xy = 0^2

29-2×2 = 0

29-4 = 0

= 25