Anyone who is really passionate about maths?

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Aww...you haven't scared me at all. I understand how you feel. Remember, to be a success in life, you need to find out what you love to do, and find a way to make that a career. You're right; not all can be math geniuses (like VectorSpace), but if you find out what you love to do, and make that a career, you will find happiness and be able to spread that to others. I hope you do well in whatever you do.

I don't think you did anything bad. OK, you were given the solution, but you wanted to understand it.

Ok I'm working on a third grade equation now. I know this is probably a long shot but, could I not just make it a second or ordinary equation and solve it like that?

Ok I get it. Thank you!

This feels wrong to me but I have no idea what I might have done wrong.

[img][800:576]http://farm4.staticflickr.com/3821/9205221636_c820a37c38_b.jpg[/img]

Hi, Salome. I hope you're doing well. I'm afraid that you have made three errors in this problem.

The first is that instead of dividing by x (even though all three terms had an x in common), you need to factor out the x so that you have this:

Note: for notational purposes, I will use the ^ sign to represent an exponent...like x^2 means x squared, and x^3 means x cubed.

The original problem then would look like this:

x^3 - 4x^2 - 5x = 0 (confusing, isn't it? Well, thanks to Bill Gates for not providing us with a good math text editor that can work anywhere instead of just Microsoft Word...heh)

Now, since all three terms have an x in common, we can factor out an x so it would now look like this:

x(x^2 - 4x - 5) = 0. Why the &$% would you do that? Because x = 0 is one solution to that equation. Look -> 0(0^2 - 4(0) - 5) = 0.

When you divided that whole thing by x, you eliminated one of the solutions. But please don't leave x= 0 in the cold cold snow! He has a wife and family! Why, everyday he has to walk 10 miles to the bus barefoot in the snow! Just kidding around...

The second mistake is that on the third line down on your step by step written out page, you have to follow the ordering of stuff. You need to divide 4 by 2 first before you square it under the square root. Then you'll get 2 squared, which is 4.

But wait, there's more (shuts off the TV because of the salesman's annoying voice)...

Underneath that radical, you will then have 4 + 5. But that equals 9, and the square root of 9 is 3. So now you should have this:

x = 0 (don't forget him), x = 2 +/- (that's plus or minus) 3. This means that your answer is 0, -1, and 5.

What? Is it possible to have three solutions? Of course it is. It is a cubic equation after all, and the most solutions you can have is 3. (jokingly) 3 is the number that you shalt count, and the number that you shalt count is 3...

I only hope that this helps you, Salome. If not, please let me know how I can help. Have a great evening.

By the way, if you just want to check if your solution is right or if you want to know what the function "looks like", Wolfram Alpha is really helpful:
[url]http://www.wolframalpha.com/input/?i=x^3-4x^2-5x[/url]

I made the mistake to use a similar program for my homework in high-school because I was bored by the assignments, but it resulted in lots of arithmetical errors in the exams.