Numbers

[OhMissScarlett]

Omg, numbers. *hides*

p.s. I just tried to call you and obviously you're not home.

 

[carsonshepherd]

Omg, numbers. *hides*

p.s. I just tried to call you and obviously you're not home.

Oh shit, yes I am! *jinlges you*

 

[3113]

6x(x+1)(3x+3)-x(5x)(3x+3)=5x(x+1)

6x(x+1)(3x+3)-x(5x)(3x+3)-5x(x+1)=0

6x(x+1)(3x+3)-x(5x)(3)(x+1)-5x(x+1)=0

With you up to this point. But how did you get rid of all three of the (x + 1) to get to this step?

6x(3x+3)-x(5x)(3)-5x=0

 

[ImpishSeraph]

a) -13/15 b)-13/3 c) -13 d) no solution

I get b) -13/3 when i work it out.

I factored out the 3 on the second part of the left side and then x-1 will cancel out. I ended up with 18x^2 +18x -15x^2 = 5x. Combine like terms and you get 3x^2 + 18x = 5x. Then factor out 3x on the left side and get 3x(x+6) = 5x. Divide both sides by 3x and get x+6 = 5/3. At this point when you subtract 6 from both sides you get -13/3 which is b. You can plug it back in and check to see that it works.

 

[BlackShanglan]

I have the opposite problem as you. I have no problem with the concepts, I just don't do arithmetic well.

That's me. On every math test I've ever taken, there's been some impressively complex problem that I got completely right except for, at some critical juncture, managing to add 3 and 4 and come up with 8.

I also cannot remember numbers. Not even for a few seconds. I have to chant them over and over if I want to recall them while I walk from one room to the next. 144 lines of junk poetry I memorized when I was 14, however, is still safely locked in memory.

 

[Ǽltya]

With you up to this point. But how did you get rid of all three of the (x + 1) to get to this step?

6x(x+1)(3x+3)-x(5x)(3)(x+1)-5x(x+1)=0

From this step you cancel out the (x+1) from each section by diving them all by (x+1). Zero divided by anything remains zero so the right side doesn't change.

6x(x+1)(3x+3)-x(5x)(3)(x+1)-5x(x+1)=0

 

[carsonshepherd]

I still have to divide imaginary numbers....

 

[3113]

6x(x+1)(3x+3)-x(5x)(3)(x+1)-5x(x+1)=0

From this step you cancel out the (x+1) from each section by diving them all by (x+1). Zero divided by anything remains zero so the right side doesn't change.

Got it, but....

I still have to divide imaginary numbers....

I'm in agreement. This is the part I could never get, deciding which number to cut out by doing anything (like subtracting or multiplying or dividing). In this case, as you can divide by anything (As zero remains zero), why go for the (x + 1). Why not the 5x? Or the (3x + 3)? How do you decide, out of the blue, I'm gonna get rid of those (x + 1)'s and here's how I'm gonna do it?

Or is this like asking a writer, "How do you come up with your ideas?" Do you number geeks just know? And if so, then can the rest of us ever be taught this superpower?

 

[JAMESBJOHNSON]

Math is useful. I used it all the time as an engineer. Solving for unknowns saves tons of time over trial & error.

Geometry is my love, though. One of my professors gave us a geometry problem that took me 20 years to solve because its a paradox; that is, the surface area of the puzzle increases or decreases depending on how you arrange the pieces. The professor had no explanation for it.

The problem with most math teachers is they cannot explain what they do. The process is outside of their conscious awareness. Like riding a bike. Or walking. So many math students try to follow the teacher's example and fall over on their ass the first time they mount the bicycle or take a step.

 

[trysail]

6x(x+1)(3x+3)-x(5x)(3x+3)=5x(x+1)
6x(x+1)(3x+3)-x(5x)(3x+3)-5x(x+1)=0
6x(x+1)(3x+3)-x(5x)(3)(x+1)-5x(x+1)=0
6x(3x+3)-x(5x)(3)-5x=0
18x2+18x-15x2-5x=0
3x2+13x=0
3x+13=0
x=-13/3

[exponents in red font]
I'm such a frickin' geek.

Bravo!

"Anyone who cannot cope with mathematics is not fully human. At best, he is a tolerable subhuman who has learned to wear shoes, bathe, and not make messes in the house."

-Robert A. Heinlein
Time Enough For Love
"The Notebooks of Lazarus Long"

 

[Pure]

dear, aelt,

you can't just discard factors!

it's initially a cubic equation, and has, thererefore, three solutions (as an equation, apart from the practical constraints*).

x=0, x= -1, and x= -13/3

===

 

[carsonshepherd]

Got it, but....


I'm in agreement. This is the part I could never get, deciding which number to cut out by doing anything (like subtracting or multiplying or dividing). In this case, as you can divide by anything (As zero remains zero), why go for the (x + 1). Why not the 5x? Or the (3x + 3)? How do you decide, out of the blue, I'm gonna get rid of those (x + 1)'s and here's how I'm gonna do it?

Or is this like asking a writer, "How do you come up with your ideas?" Do you number geeks just know? And if so, then can the rest of us ever be taught this superpower?

Because ridding the equation of the (x+1) makes it a binomial. Ridding the equation of the other numbers doesn't do you any good.
I think it can be taught, because I'm not a number geek, but I can follow rules and that's all math is - a set of rules.

 

[carsonshepherd]

you can't just discard factors!

it's initially a cubic equation, and has, thererefore, three solutions (as an equation, apart from the practical constraints*).

x=0, x= -1, and x= -13/3

===

Yes you can, when you divide by a factor. That makes it a binomial, with one real solution.

 

[impressive]

I think it can be taught, because I'm not a number geek, but I can follow rules and that's all math is - a set of rules.

Until you get to theoretical calculus. *shudder* That one damned near prevented my graduation.

 

[carsonshepherd]

Until you get to theoretical calculus. *shudder* That one damned near prevented my graduation.

*laughs* there's no worry that I will EVER get to any kind of calculus.

 

[Misty_Morning]

6x(x+1)(3x+3) - x(5x)(3x+3) = 5x(x+1)

3x+3[(6x)(x+1) - (x)(5x)] = 5x(x+1)

3(x+1)[(6x)(x+1) - (x)(5x)] =5x(x+1)

now you can divide each side by (x+1):

3[(6x)(x+1) -(x)(5x)] = 5x

Now you can divide each side by x:
3[(6)(x+1) - (5x)] = 5
3[6x+6-5x] = 5
18x +18 -15x = 5
18x-15x = 5-18
3x= -13
x=-13/3

 

[neonlyte]

I love it when you all talk dirty

 

[Pure]

note to carson.

Originally Posted by Pure
//you can't just discard factors!

it's initially a cubic equation, and has, thererefore, three solutions (as an equation, apart from the practical constraints*).

x=0, x= -1, and x= -13/3//

===
Carson:

Yes you can, when you divide by a factor. That makes it a binomial**, with one real solution.

Unfortunately, you are not correct. There are three real solutions to the original equation.

If you don't agree, please tell me how many real solutions this equation has:

x(x+1)=0 , alternatively written:

x**2 + x = 0

==
**NOTE: the first division by the factor (x+1) did NOT produce a binomial.

===

note to misty

now you can divide each side by (x+1)

no, you can't without altering the situation. for instance, suppose x=-1.

there are certain operations that cannot unthinkingly be done to both sides of an equation, e.g. taking a (positive) square root.

for example x**2=4.

nor can you square each side, transforming x=2 into x**2=4

 

[Misty_Morning]

Originally Posted by Pure
//you can't just discard factors!

it's initially a cubic equation, and has, thererefore, three solutions (as an equation, apart from the practical constraints*).

x=0, x= -1, and x= -13/3//

===
Carson:

Yes you can, when you divide by a factor. That makes it a binomial**, with one real solution.

Unfortunately, you are not correct. There are three real solutions to the original equation.

If you don't agree, please tell me how many real solutions this equation has:

x(x+1)=0 , alternatively written:

x**2 + x = 0

==
**NOTE: the first division by the factor (x+1) did NOT produce a binomial.

===

note to misty

now you can divide each side by (x+1)

no, you can't without altering the situation. for instance, suppose x=-1.

there are certain operations that cannot unthinkingly be done to both sides of an equation, e.g. taking a (positive) square root.

for example x**2=4.

nor can you square each side, transforming x=2 into x**2=4

I beg to differ....

you can divide both sides of an equation by any number or variable... you have to perform the same function on both sides.

the name of the game is called factoring out common variables and then getting rid of those common varibles in order to simplify the equation.