I. V.

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Quadratic equation x²+[a²-5a+b+4]x+b=0 has roots -5 and 1,then nuber of integral values a is:(where[.] denotes the greates integer function)

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2 Answers

Aniket Ghosh Dastidar ·2011-03-09 23:17:30

product of roots =b=-5
therefore the term in [.] becomes [a²-5a-1]
now sum of roots is -4
therefore -[a²-5a-1]=-4
or [a²-5a-1]=4
therefore a²-5a-1 must lie in [4,5)
Now take both the cases once when it is equal to 4 and when it is equal to 5..
equating it with 4 roots are -0.85 and 5.85
equating it with 5 roots are -1 and 6..
so the solution set is (-1,-.85]U[5.85,6)..
Since there is no integral value in this range...so answer is 0.