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boraxmatix · May 22, 2009, 10:19 PM

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

galactus · May 23, 2009, 11:01 AM

Originally Posted by boraxmatix

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]

boraxmatix · May 24, 2009, 12:24 AM

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

Perito · May 24, 2009, 04:27 AM

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 $(-\infty, \;\ -20)$

boraxmatix · May 25, 2009, 12:31 AM

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

boraxmatix · May 25, 2009, 12:34 AM

Now will the same logic hold for the inequality

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

Perito · May 25, 2009, 05:16 AM

Yes, absolutely.