Toribash
Original Post
[Event] ynvaser's Maths Challenge

Hey there, guys! Since I have some TCs and items to burn, I've decided to spend it on a little event.

Uncle Dave? What is this sh&#?
I'll post a math problem. If you solve it, and provide a step-by-step guide on how you solved it, I'll give you the amount of TCs plus an item specified below. Cheating (using a program to do it for you) will be apparent to me, and you'll be disqualified. You can either PM me your solution, or if you are feeling brave, you can post it here.

That sounds great! Show me the problem!
Our first problem will be the following equation. I'd like to know the real roots of it (x€R).
2x^4-5x^3-8x^2-5x+2=0

Neat! The prize?
2000 TC +

Remember, your answer is valid only if you provide a step-by-step solution! Have fun, lads
Last edited by ynvaser; Dec 17, 2012 at 05:43 AM.
Originally Posted by mclarensf1 View Post
Heres go nothin. Let's hope Honors Math helps. xD

Umm, nope. 5x^3 isn't 125x, for example, it's 5x*x*x

Originally Posted by torikill10 View Post
Fuck. Can't do this. Now you think that I suck. But I haven't learned this. Maybe next year. Just for explaining.

This is high school maths. So maybe then.
Originally Posted by ynvaser View Post
Umm, nope. 5x^3 isn't 125x, for example, it's 5x*x*x

That explains a lot. I kept wondering if the ^3 applied only to the variable or to the number too.
If it's (abcdx)^3, then it's (a^3)*(b^3)*...*(x^3). If it's abcdx^3, then it means a*b*c*d*(x^3). Rising to a power is a higher level operator than division and multiplication.
i don't even know what i'm doing but w/e

2x^4-5x^3-8x^2-5x+2=0

x^2(2x^2-5x-8 )=0+5

2x^2-5x-8=5

2x^2-5x-13=0

(5+|- √ √-5^2-4*2*-13)/4

(5+|- √129)/4

x=4.09 ^ x=-1.59
Last edited by pusga; Jan 22, 2013 at 09:30 PM.
Wolfram is right, and it's doable. I'll show you how:

Let's divide by x^2 (we'll lose 2 imaginary roots this way, but we don't need those anyway).

Let's rearrange the equation a bit, so we get the following:

Now let's say that
Then y^2 would be:
That looks quite similar to the first part of our equation, we just need to subtract 2 and we'll have what we need, which looks like this:

Now, we have a plain quadratic. Solve it using the quadratic formula, then substitute back to get the two roots of x we were looking for (2+-square root 3). Check it yourself if you don't believe me :P Event is over.