The Annoying Math Game

tony_montana

tony_montana

Semper Fidelis
I don't know why Im making this game up. Im just bored and this will give us something to do!:D Plus I think we could use another game.
Anyway, the game is simple. Im gonna put a math problem here(annoying algebra type math:D ) and see who can answer it. Whoever answers it gets to make up the next problem. Simple, eh?

Here it goes!
2(x+2)=?
And that's the ANNOYING MATH PROBLEM!:D
 
Y

yardgames

Retired Administrator
Here's a much harder one. Tell me the vertex of the graph of the following equation: y = -5x^2 + 20x -3 (note: that's 5 x squared)
 
D

Dabney

Deutscher Moderator
@yardgames: May the vertex be (-2/-63)? I didn't try too hard, so probably it's wrong...


and @J.: For your solution x=-2 the equation would have to be 2(x+2)=0, but Mike's one was only 2(x+2)=?, so I would have stopped calculating with 2x+4.

Does that make any sense or am I talking total crap?
 
NeCoHo

NeCoHo

Retired Mod
It was asked to do both, as in both finding the variable, and solving it, unless I misinterpreted it...
 
admin

admin

Retired Administrator
I haven't done maths for years now, but I always thought you needed simultaneous equations to calculate 2 independant variables. Currently that equation is possible with an infinite amount of numbers because ? will always be equal to 2x + 4. Therefore you can't have a wrong answer.
 
Y

yardgames

Retired Administrator
@Dabney: No, that's not correct. Sorry.

If it helps, you don't actually have to graph it. It's much more precise to solve the equation. Which of course you can't do because you have an x squared and an x, which aren't like terms and cannot be combined. So you need to convert the equation to something usable. I recommend using a perfect square trinomial.
 
tony_montana

tony_montana

Semper Fidelis
Ok, who solved the problem first? Yardgames, J, or someone else?(id like to do a turn based thing. whoever solves gets to make the next problem and they can make it extremely easy or extremely hard!:D )
 
Wildcat

Wildcat

Retired Moderator
y = -5x^2 + 20x -3

Basic quadradic equation: y = ax^2 + bx + c

Formula for the axis of symmetry: x = -b / 2a

Solving for x:

x = -20 / 2(-5)
x = -20 / -10
x = 2

Solving for y:

y = -5 (2)^2 + 20 (2) -3
-5 (4) + 40 -3
-20 + 40 - 3
y =17

Vertex = (2, 17)
 
Y

yardgames

Retired Administrator
Yep, that's right. Damn, I should have asked you when I was trying to figure out how the hell to do this stuff-that's a lot easier than the way my teacher taught me. :D Of coures, after he taught us, he gave us a calcuator program that does it automatically--it's great, just type in the numbers and it finds x-intercept, y-intercept, vertex, and a bunch of other crap for you.
 
Wildcat

Wildcat

Retired Moderator
yardgames said:
Yep, that's right. Damn, I should have asked you when I was trying to figure out how the hell to do this stuff-that's a lot easier than the way my teacher taught me. :D Of coures, after he taught us, he gave us a calcuator program that does it automatically--it's great, just type in the numbers and it finds x-intercept, y-intercept, vertex, and a bunch of other crap for you.
Click to expand...

Yeah, I was wondering why you suggested taking the perfect square trinomial. I guess you could do it that way, but it seems like it would be a lot harder and more and more involved. I hate it when they make you do all that work then tell you that you could just plug it into the calculator. They just do it to torture us. :D

I'll let someone else post the next one cause I got rid of my algebra book and I can't think of anything off the top of my head.
 
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