Question

Hi, I'm having trouble with college agebra
can anyone help?

Answer 1

any specific question in mind?

Answer 2

I'm currently studying linear algebra, ask away.

Answer 3

Yes. Actually I have a BIG final tomorow and I'm very lost. I was wondering if anyone who is good at college algebra would mind helping me with a few problems. One is a quadratic problem..... it's f(x)=4x^2+40x+9

Answer 4

I have the anwers...it is (-5,-91) my question is HOW? I'm not sure where to start on this problem

Answer 5

Sure. The best and easiest way to solve quadratics is to throw it into the quadratic formula. You should always do this before you try any other methods of solving it.

Do you know the quadratic formula?

Answer 6

x^2=k??

Answer 7

http://images.wikia.com/algebra/images/6/61/Quadratic_Formula.gif

Answer 8

Where it follows ax^2 + bx + c

Answer 9

So basically what you want to do in these kinds of questions is to just substitute the coefficients into the quadratic formula. Solve it for both the + and the -, and you will get your two roots.

Answer 10

Is there some kind of homeowrk hotline number I can call that will allow me to talk to someone while I'm solving on here? This sounds silly but I have to pass this test tomorrow to pass the class

Answer 11

Aha. Do you have Skype? I'll be more than happy to walk you through it.

Answer 12

I can download it

Answer 13

Practice is the best way to study, though, in the long run. If you want, you can try some problems, and if you get stuck, feel free to post them here. There are a ton of people who can help.

Answer 14

okay here is another.....2x^2-4x+5=0....keep in mind I have the answers to the problems. I am just looking for an easy HELPFUl way to solve these. I need things broken dow

Answer 15

Ok, sure no problem. Btw, if you want more publicity for your call for help you should post a new question.

give me a sec for the problem

Answer 16

Okay thank you. I am new at this site

Answer 17

no problem. Anyway, back to math:
for all quadratic equations, it's nice to use the quadratic formula if you don't know how to factor, or if factoring doesn't work. So there are at least 2 ways to solve this:

Quadratic Formula Method:\[{-b\pm \sqrt{b^2-4ac} \over 2a} = {-(-4)\pm \sqrt{(-4)^2-4(2)(5)} \over 2(2)}\]

Answer 18

Then you simplify and you should get \[\pm \sqrt{-24}\]
Since the number under the radical is negative, your answer is not real. IDK if you've learned imaginary, but if you did, you can factor the radical to \[\sqrt{-1*4*6} = 2i\sqrt{6}\]

Answer 19

And factoring doesn't work here, btw