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If the equations x²-10cx-11d=0 are a,b and those of x²-10ax-11b=0 are c,d then value of a+b+c+d , is (where a,b,c and d are distinct number)
a2 - 10acx - 11d = 0 c2 - 10acx - 11b = 0 subtract them ...... a2 - c2 = 11 ( d - b ) ........ (1) a + b = 10c, ab = - 11d, c + d = 10a, cd = - 11b a + b + c + d = 10 ( a + c ) ...... (2) b + d = 9 ( a + c ) ....... (3) Also, abcd = 121bd so ac = 121 from result (1) & (3) ..... ( b + d ) ( a - c ) / 9 = 11 ( d - b ) ( a - c ) / 99 = ( d - b ) / ( b + d ) Simplifying. ........... a2 + c2 - 20 ac - 11 ( b + d ) = 0 ( a + c )2 - 22ac - 11 ( a + c ) 9 = 0 substutuing dat ac = 121 u get finally (a+c)2 - 99 (a+c) - 2662 = 0 This is a quadratic in a+c solving u get a + c = 121 Go to result ::: a + b + c + d = 10 ( a + c ) = 1210
a2 - 10acx - 11d = 0
c2 - 10acx - 11b = 0
subtract them ......
a2 - c2 = 11 ( d - b ) ........ (1)
a + b = 10c, ab = - 11d,
c + d = 10a, cd = - 11b
a + b + c + d = 10 ( a + c ) ...... (2)
b + d = 9 ( a + c ) ....... (3)
Also, abcd = 121bd
so ac = 121
from result (1) & (3) .....
( b + d ) ( a - c ) / 9 = 11 ( d - b )
( a - c ) / 99 = ( d - b ) / ( b + d )
Simplifying. ...........
a2 + c2 - 20 ac - 11 ( b + d ) = 0
( a + c )2 - 22ac - 11 ( a + c ) 9 = 0
substutuing dat ac = 121
u get
finally
(a+c)2 - 99 (a+c) - 2662 = 0
This is a quadratic in a+c
solving u get
a + c = 121
Go to result :::
a + b + c + d = 10 ( a + c ) = 1210
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