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Question 961985: If x is an integer such that x^2-3x<4, then the no. Of possible values of x is plz describe that ans.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! to solve this problem, you need to find the zero points and then test the equation within each of the intervals established by those zero points.
your equation is x^2 - 3x < 4
subtract 4 from both sides of the equation to get x^2 - 3x - 4 < 0
set the equation equal to 0 and solve for x.
x^2 - 3x - 4 = 0 is a quadratic equation.
factor this equation to get (x-4) * (x+1) = 0
solve for x to get x = 4 or x = -1
those are your zero points.
test the original equation within the intervals established by those zero points.
you will find that the equation is false when x is less than -1 and when x is greater than 4.
you will find that the equation is true when x is between -1 and 4.
that's your solution.
since x has to be an integer, then the number of possible values between -1 and 4 is 4.
those numbers are 0,1,2,3
here's a graph of your equation of x^2 - 3x - 4 < 0
note that you could also have graphed two equations separately.
the first equation would be x^2 - 3x
the second equation would be 4
that graph would look like this:
the intersection points are the same in both graphs.
with x^2 - 3x - 4 < 0, the intersection points are (-1,0) and (4,0).
with y = x^2 - 3x and y = 4, the intersection points are (-1,4) and (4,4).
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