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Mathematics 18 Online
OpenStudy (anonymous):

Solve for the inequality: 5(x^2-1)>24x

OpenStudy (anonymous):

5x^2 -24x -5 is the first step i think

OpenStudy (anonymous):

the y intercept would be -5

OpenStudy (e.mccormick):

\[5(x^2-1)>24x \implies \\ 5x^2-5>24x \implies \\ 5x^2-24x-5>0 \]

OpenStudy (anonymous):

i understand that part...its the factoring for the x-intercept I don't understand how to do

OpenStudy (e.mccormick):

So you had it mostly right, you just need to keep the >0 in there. Then solve fr the 0s, like you would a normal factor the polynomeal. However, the lines you get are NOT part of the solution. Why? Because >, not \(\ge\).

OpenStudy (e.mccormick):

Your factors for 5 are 5 and 1 because it is prime. That is nice because you have a -24 in the middle, so you know you need \(5\times (-5) +1 = -24\) involved in there somewhere.

OpenStudy (e.mccormick):

\(5x^2\) can only factor into \(5x\) and \(x\). \(-5\) can only factor to \(-5\) and \(1\) or \(5\) and \(-1\). So it must be some combination of those. The combination has to add to \(-24\) Well, \(5\times (-5) +1 = -24\) So I MUST have: \(5x^2\) factored into \(5x\) and \(x\). \(-5\) factored into \(-5\) and \(1\). What way do you think that would come apart as factors? \((\,?\,\pm\,?\,)(\,?\,\pm\,?\,)\)

OpenStudy (amistre64):

if your not asked to factor, but just to determine the x intercepts, a good method would just to employ the quadratic formula .... but factoring is fine if you can find rational solutions

OpenStudy (anonymous):

\[(5x \pm x) (-5 \pm 1) \] ?

OpenStudy (e.mccormick):

No, it needs to be able to FOIL, or multiply back into the equation. I wish I had a good reference on factoring in German I could point out to you. The factor will start with this: \((5x\pm \,?\,)(x\pm \,?)\). That way those two will multiply back to the \(5x^2\). This means the \(\pm\,?\) must be the \(+1\) and \(-5\), but you still need to put them in the right place to get that \(-24x\) term in the middle.

OpenStudy (anonymous):

I am not German haha so an English reference would suffice

OpenStudy (e.mccormick):

Oh, just put German on your profile page. Hehe. Tricked me!

OpenStudy (anonymous):

I speak it lol but I am American

OpenStudy (anonymous):

So, to find the order should I guess and FOIL it...and if it is wrong try the other combo?

OpenStudy (e.mccormick):

I have not spoken much German in years, but I surprised a customer with a little German and she is amazed at my accent. I learned French and English as a child, so getting accents right is easy. Now back to math. The guess and try method works. The goal is top get better at understanding it so the guesses are better each time. There are also some tactics, like how I decided is must be -5 and 1.... which should point with a huge arrow to what needs to be multiplied. The -24 is the major clue there. http://www.purplemath.com/modules/factquad.htm

OpenStudy (anonymous):

ok would it be \[(5x \pm -1) (x \pm-5)\]

OpenStudy (e.mccormick):

Well, you can kill the \(\pm\) because you have the signs now. Otherwis, exactly right for that part.

OpenStudy (anonymous):

oh sorry, so (5x -1) and (x - 5)?

OpenStudy (e.mccormick):

Oh, and +1!!!! Not-1!

OpenStudy (anonymous):

OH that makes sense lol

OpenStudy (e.mccormick):

With what we talked about so far, you ended up with \((5x+1)(x-5)>0\). But that is not the answer. It just helps us get close to the answer. I have worked up a bit of a thing on the rest, I can post it, or I can keep giving hints and leting you work it.

OpenStudy (anonymous):

If you could please keep giving me hints, if I dont learn I wont pass the test on it

OpenStudy (e.mccormick):

Goog, good. OK. We are at \((5x+1)(x-5)>0\) The next step to solbing this is where is this true? Now, we know that the \(>\0\) part means that anything that imakes the left 0 is not tue, but it is the borderline between what is true and what is not true. Because we have the left factored, you can solve each factor for 0.

OpenStudy (anonymous):

so 5x+1=0 5x=1 x=1/5 and x-5=0 x=5 and then find the vertex?

OpenStudy (e.mccormick):

Well, no a vertex. But that part is correct. Next is test points. Here is that setion of what I wrote: Lets look at a couple of things here. \((5x+1)(x-5)>0\) means that if I put in \(-\frac{1}{5}\) I get: \[(5(-\frac{1}{5})+1)(x-5)>0\implies (-1+1)(x-5)>0 \implies (0)(x-5)>0 \implies 0>0 \]And that is just not true! So \(-\frac{1}{5}\) is a critical point here. Similarly, the \((x-5)\) part means 5 is another critical point. Once you know these critical points, you need to test around them. So you have everything below \(-\frac{1}{5}\) to test, everything between \(-\frac{1}{5}\) and 5 to test, and everything above 5 to test. The good news is you only need one test for each.

OpenStudy (e.mccormick):

So, what test points would you pick? I go for simple numbers, ones that are close, and thigns like that.

OpenStudy (e.mccormick):

Your book might call these a truth table. You test where things are true or false and that helps define the answer.

OpenStudy (anonymous):

I dont know how to do one lol

OpenStudy (e.mccormick):

Fine, fine. We can just look at these points on a line.|dw:1366836933109:dw| This is both the line and the table. THe line has the marks and my bad curves are the areas we want to check, abovem between, and below. The table version is in the lower left.

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