The set of values of "a" for which one root of x^2 + ax + a + 3 = 0 is greater than 1 and the other less than 1 is (a) (6,infinity) (b) (-2,6) (c) (-infinity,-2) (d) (2,infinity)

Asked by buttercup | 22nd May, 2012, 11:17: PM

Expert Answer:

We solve we use hit and trail,
For (a) (6, infinity)
take, a = 7
equation is: x2 + 7x + 10 = 0
this gives, x = -5 and -2, not valid
 
For (b) (-2,6)
take a = 0
equation is: x2 + 3 = 0
 this gives complex roots, so not valid
 
For (c) (-infinity, -2)
take a = -3
equation becomes, x2 - 3x = 0
this gives, x = 0 and 3
That is one root is greater than 1 and other is less than 1.
Hence this is the correct option.

Answered by  | 8th Jun, 2012, 05:41: PM

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