Hi
How do we solve an inequality like (x-6) (x +7) > 0


or (y+1) (y) (y-2)(y-4) greater than or equal to 0

thanks in advance

with due regards

Shekhar

Hi
How do we solve an inequality like (x-6) (x +7) > 0

(-\infty, \;\ -7)\cup (6, \;\ \infty)


or
(y+1) (y) (y-2)(y-4)\leq 0


(-\infty, \;\ -1]\cup [0, \;\ 2]\cup [4, \;\ \infty)


thanks in advance

with due regards

Shekhar[/QUOTE]

But how did u get to the answer... that is what I want to know... thanks

How do we solve an inequality like (x-6) (x +7) \gt 0

(y+1) (y) (y-2)(y-4) \ge 0


It's just like solving the individual equations. Don't let the inequalities bug you too much. For example,

(x-6) (x +7) \gt 0

This is already factored. It's not like x^2+x-42 \gt 0, which is the same equation unfactored. Since it's already factored, a lot of your work is taken away. Treat it like this equation:

(x-6) (x +7) = 0

You know that if the product of two terms is equal to zero, one of the terms must be equal to zero. Therefore

x-6=0 and x=6

or

x+7=0 and x=-7

Now, since this is an inequality, you put the two answers together using set notation recognizing the boundaries imposed by the inequality.

(-\infty, \;\ -7)\cup (6, \;\ \infty)

You do the same with the other equation whose roots of the equality are -1, 0, 2, and 4. Remember that this equation includes the equality (≥ not just >).

In some cases, you can simplify the set terminology. For example, if you find that
(-\infty, \;\ -10)\cup (-\infty, \;\ -20), it's easy to show that the second set is entirely enclosed in the first set, so the union of the two sets is simply the second set (-\infty, \;\ -20)

Thanks a lot to u both... thx a lot

Now will the same logic hold for the inequality

(x- 5)(x+8)(x - 3) < 0??

Yes, absolutely.