
-----------------------------------
Martin
Wed Jul 28, 2004 11:59 am

Math Question
-----------------------------------
This one seems really easy, but my answer doesn't match the one in the back of the book. Yeah, Mr. White would kill me for saying that (and probably for asking this question too)

Anyway, here it is:
ax^3 + bx^2 - x

If divided by x - 1, remainder is 7
If divided by x + 3, remainder is 33.

Find a and b.

-----------------------------------
Martin
Wed Jul 28, 2004 12:03 pm


-----------------------------------
Nevermind. Seems my problem is that I can't do synthetic division...

-----------------------------------
Martin
Wed Aug 04, 2004 1:35 pm


-----------------------------------
New question. Why am I so stupid lately?

(2x -1) / (x -4) > 1, where x != 4

Solve for x

-----------------------------------
AsianSensation
Wed Aug 04, 2004 1:49 pm


-----------------------------------
x < -3 and x > 4

-----------------------------------
Martin
Wed Aug 04, 2004 1:56 pm


-----------------------------------
Yeah, I got it. MBB (Magic bulletin board).

I forgot to switch the inequality sign when I multiplied. The summer has been too long already

-----------------------------------
Martin
Wed Aug 04, 2004 1:56 pm


-----------------------------------
Now here's a strange one....

For f(x) = x^2 +kx +k, determine all values of k such that f(x) > 0

-----------------------------------
AsianSensation
Wed Aug 04, 2004 2:06 pm


-----------------------------------
I think it's 4 > k > 0

-----------------------------------
Martin
Wed Aug 04, 2004 2:18 pm


-----------------------------------
How do you get that?

-----------------------------------
Martin
Wed Aug 04, 2004 2:20 pm


-----------------------------------
I got the x < 4 by setting the equation to having two equal roots. Now, how do I check to see if an equation has no real roots?

EDIT: Nevermind. Quadratic equation. MBB. w00t.

-----------------------------------
Maverick
Wed Aug 04, 2004 3:24 pm


-----------------------------------
I have a question. Why are u doing math in the summer?

-----------------------------------
Martin
Wed Aug 04, 2004 3:33 pm


-----------------------------------
I like math. And I didn't do any math during highschool, so I figured that, seeing as I'm going for a math degree from waterloo, now would be a good time to start.

-----------------------------------
Maverick
Wed Aug 04, 2004 3:38 pm


-----------------------------------
lol ur going for a math degree when u never took math in HS????

-----------------------------------
Genesis
Wed Aug 04, 2004 8:13 pm


-----------------------------------
I think he meant that he didn't do any work in math during highschool. Like me. It's easy to get good math marks without really doing the homework/assignments.

-----------------------------------
Martin
Wed Aug 04, 2004 11:46 pm


-----------------------------------
Good call Genesis ;)

-----------------------------------
Martin
Fri Aug 06, 2004 10:53 pm


-----------------------------------
Okay, here comes the next question:

If the roots of the equation x^3 -15x^2 +cx -105 = 0 are a - b, a, and a + b, determine a, b, and c.

-----------------------------------
AsianSensation
Sat Aug 07, 2004 10:53 am


-----------------------------------
equation is : x^3 - 15x^2 + 71x - 105

which factors to (x-3)(x-5)(x-7)

so the roots are 3, 5, and 7

let 5 be a, then b is 2.

so a = 5, b = 2, c = 71

I think.

-----------------------------------
Martin
Sat Aug 07, 2004 3:23 pm


-----------------------------------
Yeah, it works. Does that come from T&E synthetic division, or is there a better method?

How did you get the c = 71?

b can also be -2 though.

-----------------------------------
AsianSensation
Sat Aug 07, 2004 4:39 pm


-----------------------------------
well, I just saw 105, prime factored, it, you get 3, 5, 7. That really looks like the a-b, a, a+b format. Multiply it out, you get that, and then just answer the question from there.

-----------------------------------
Martin
Sat Aug 07, 2004 4:41 pm


-----------------------------------
Good call. I'll look for that in the future. Thanks a ton for all of your help by the way :D:D

-----------------------------------
AsianSensation
Sat Aug 07, 2004 5:46 pm


-----------------------------------
no problem, I'll just get the favor back by asking compsci help from you sometimes.

I needed to brush on math anyways, I'm taking the SAT, again..........ghey.
