Math Is Fun Forum

shocamefromebay

Member

Registered: 2007-05-30

Posts: 103

Inequalities and negative numbers?

I am having trouble with an inequalities problem.

I found the solution to be

, but the real solution is

.

Why is that? Is there some sort of rule that I do not know about? does it have somethign to do witht eh variable being in the denominator?

TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: Inequalities and negative numbers?

I'd guess these were your steps:

The key here is moving from step 2 to step 3.  You're multiplying by 8a, which is positive when a is positive and negative when a is negative.  Whenever you multiply or divide both sides of an inequality by a negative number you need to switch the sign around.

To show you what would happen we'll split the problem into 2 sections: first is where a > 0, the second is where a < 0 (it should be obvious what happens when a = 0).  If we assume a > 0 then we leave the sign as it is and can use the proof I gave above.  Let's see what happens if we assume a < 0:

Thus we've found that a > 7/4.  However, we assumed that a < 0.  This is a contradiction, which means that there is no solution when a < 0.  You can see that a = 0 is not a solution either, so we're left with 0 < a < 7/4.

---

Wrap it in bacon

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Inequalities and negative numbers?

An alternative method would have been to multiply by 8a², which is always positive and so means you don't need to split into cases. You'd then get to 4a² - 7a < 0, which is a quadratic that solves to give the same solution.

---

Why did the vector cross the road?
It wanted to be normal.

shocamefromebay

Member

Registered: 2007-05-30

Posts: 103

Re: Inequalities and negative numbers?

Wow, thanks guys. It was a real help!