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![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 Mar 2007 15:15:21 IST
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a(x2-y2) +  [x(y+1) +1] can be resolved into linear factors . Then:
(a) =1
(b) = 4a2/(a-1) [a 1]
(c) =0,a=1
[ans: (b),(c)]
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3 Mar 2007 15:16:24 IST
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4 missing objects on left are lambda and (b)....[a is not equal to 1].
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3 Mar 2007 15:16:42 IST
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what are the missing links?
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-Ramya |
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7 Mar 2007 18:03:03 IST
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R u sure you have keyed in the correct question...
Because the condition for havin two linear factors of the general eqn.
ax2 + by2 + 2gx + 2fy + 2hxy + c = 0 is
abc + 2fgh - af2 - bg2 - ch2 = 0
Using this we get...
L(L2 - L + 4a2) = 0
Which gives one of the answers as L = 0, but the other ans is not satisfied.
(L stands for lambda)
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Sudeep Kumar
(B tech, IITd)
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