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If the equation x²+ 3x + a =0 has no negative root, then which of the following could be the value of a?
(A) 8 (B) 7 (C) 4 (D) 9
given x2+3x+a = 0, given equation has no negative root, also b/a =3 , hence sum of roots = -3, this implies equation has two non real/ imaginary roots, 9-4a<0 -> a - (2.25, infinity)
Using hit and trial, all the options are wrong. Since, b^2 - 4ac is negative for all options, so the equation had no real and imaginary roots.
If we solve b^2 - 4ac for the given condition, then
b^2 - 4ac greater than 0
9 - 4a greater than 0
9 greater than 4a
so the possible value for a is 1 and 2
But these values still doesn't satisfy the condition of no negative roots
@utkarsh How have you arrived all the above pls explain
When sum of roots and product of roots are positive, then both roots are positive..but here sum of roots =-3 and Product of roots is=a... if "a" is positive , positive roots are zero, negative roots are 2, and "a" is -ve , positive roots are 1 and negative roots are 1...
It is contradicting given statement that there are no negative roots
