262i dont think it is as good as u hv mentioned.
simply, |x + y + 1|2 = x2+y2+1+2(x+y+xy) = 6+1+4+6√2 =(3+√2)2 .
given,
x + xy + y = 2 + 3√2 and,
x2 + y2 = 6
Then find the value of |x + y + 1|
options are :
1) 1 + √3
2) 2 - √3
3) 2 + √3
4) 3 - √2
5) 3 + √2
This is a good question no doubt...!!
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4 Answers
262
36yup but you made it simple....
the solution i saw was a bit complex sort of ....!!
21x+xy+y=2+3\sqrt{2}
\Rightarrow 2x+2xy+2y=4+6\sqrt{2}
x^2+y^2=6
\Rightarrow x^2+y^2+1=7
adding both we get,
x^2+y^2+1+2xy+2x+2y=11+6\sqrt{2}
\Rightarrow(x+y+1)^2=11+6\sqrt{2}
\Rightarrow(x+y+1)^2=(3+\sqrt{2})^2
\Rightarrow|x+y+1|=3+\sqrt{2}