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![[Post New]](/templates/default/images/icon_minipost_new.gif) 26 Mar 2007 12:46:12 IST
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If a,b,c are real nos such that a^2+2b=6, b^2+4c=-7, c^2+6a=-13
then a^2+b^2+c^2=?
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I think, therefore I am |
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26 Mar 2007 15:16:20 IST
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a^2+2b=6, b^2+4c=-7, c^2+6a=-13
add all three
a2+b2+c2+2b+4c+6a=-14
or
(a2+6a+9)+(b2+2b+1)+(c2+4c+4)=0
or
(a+3)2+(b+1)2+(c+2)2=0
so
a=-3 b=-1 c=-2
hence
a2+b2+c2=14
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26 Mar 2007 15:17:44 IST
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Add all equation togather:
a2+b2+c2+2b+4c+6a= -14
(a2+2.3a + 32)+(b2+2b+1)+(c2+2.2c+22)=0
(a+3)2+(b+1)2+(c+2)2=0
Since this is sum of squares of real numbers, hence they must be individually equzl to 0
i.e. a= -3
b= -1
c= -2
a2+b2+c2 =14 Ans
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