If x^2+y^2=29and xy=2 , find the value of x+y andx-y
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Answered by
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__
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__
Answered by
0
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
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