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![[Post New]](/templates/default/images/icon_minipost_new.gif) 14 Apr 2008 13:13:08 IST
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c+d = 10a................................(1)
a+b = 10c .............................. (2)
a+b+c+d = 10(a+c)
subtracting (1) from (2)
b-d = 11(c-a)..........
now
x^2 - 10ax -11b =0 as c is a root of equation
c^2 -10ac - 11b =0................(3)
x^2-10cx-11d =0 as a is a root of equation
a^2 - 10ac- 11d =0 ...............(4)
[(3) - (4)]
(c-a)(c+a) -11(b-d) =0
put value of (b-d) from above..........
(c-a)(c+a) -121(c-a) =0
dividing by (c-a), we have ;
c+a = 121
a+b+c+d = 10(a+c) = 10*121 = 1210
hope it helps......cheers :)
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