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[Descan]

Alright. I got another one I'm not sure of why it becomes what it does,

"2z(z-1)^4 dz"

book says
"(1/3)(z-1)^6 + (2/5)(z-1)^5 + C"

and I worked it out as
"2((z^6)/6 - (4z^5)/5 + (3z^4)/2 - (4z^3)/3 + (z^2)/2) + C"

Also this one,
"((csc^2)(4x))/cot(4x)"

becomes, book says,
"(-1/4)ln|cot(4x)| + C"

while the same different calculator gives me
"(ln(sin(4x)))/4 - (ln(cos(4x)))/4 + C"

[MagmaMcFry]

Alright. I got another one I'm not sure of why it becomes what it does,

"2z(z-1)^4 dz"

book says
"(1/3)(z-1)^6 + (2/5)(z-1)^5 + C"

and I worked it out as
"2((z^6)/6 - (4z^5)/5 + (3z^4)/2 - (4z^3)/3 + (z^2)/2) + C"

That can be easily solved using z = (z-1) + 1, so the expression becomes 2(z-1)^5+2(z-1)^4, which is really easy to integrate, because diff((z+k)^n) = n*(z+k)^n-1, so the book is correct.

What you did is just expand your polynomial and integrate the terms one by one, and that is also a correct solution. What you probably didn't notice is that both answers are the same.

Also this one,
"((csc^2)(4x))/cot(4x)"

becomes, book says,
"(-1/4)ln|cot(4x)| + C"

while the same different calculator gives me
"(ln(sin(4x)))/4 - (ln(cos(4x)))/4 + C"

Note that ln(a)-ln(b)= ln(a/b), and sin(z)/cos(z) = cot(z). Again, both answers are identical.

[inteuniso]

ffuuuuuu- why did I never post in this thread.

I've been failing in Calculus for the past 2 years. I'm okay but really depressed about my stagnant growth. WaT do.

[Descan]

Khan Academy maybe?

[Jervill]

ffuuuuuu- why did I never post in this thread.

I've been failing in Calculus for the past 2 years. I'm okay but really depressed about my stagnant growth. WaT do.

Change to a less math-intensive major?

That's what I did...a bit late, admittedly.  (I can do beginner to mid level calculus, barely, but more than that my brain shuts down.)

[3man75]

Okay so i'm doing algebra still and i've recently dropped a course in it because i was going to fail. I'm retaking it this fall but i'm starting again from the begining which i sorta remember but again MATH is so much law that it's chaotic. Kinda like real life.

Anyways i know Xto the second -9x can be grouped as (x(x-9)=0. Also know that the set answers for this equation are: x=0 and X=9 because the book told me to move the 9 to the other side.

However, what do you do when you have X to the second - 25=0?

The answer the book gives is [5,5] but i don't know how to get there. Do i divide the 25?

[Descan]

Do you mean x^2 - 25 = 0?

It's basically "What two factors multiply together to get that?" You know it's only two factors because it's only x^2, if it was x^3 or more you'd have three factors or more, depending on to what degree that ^n is.

That would be (x-5) and (x+5). You can verify that by multiplying them back together and seeing what the result is. You can remember how to multiply them by the FOIL method:
First (the two X's)
Inner (in this case, the -5 and the x from the second factor)
Outer (the x from the first factor, and the +5)
Last (the -5 and +5)

x^2 -5x +5x -25

-5x and +5x cancel, so you end up with x^2 - 25 = 0

The roots of that equation are in the factored form, (x-5)(x+5), and that results in the roots: x-5 => x = 5 (move the five over to the other side and flip the sign) and x+5 => x = -5 (again, move the five over and flip the sign)

And thus, you get [-5,5] as the roots.

Not sure why your book is saying [5,5], did you make a typo or are they both actually positive? That's wrong :V


EDIT: Also, just mathematically, you can also do "x^2 - 25 = 0 ==> x^2 = 25 ==> (square-root both sides) x = 5" but doing that loses the -5 factor, and it's wrong in terms of algebraic graphing and such-like :V. Your book might be glossing that over to not confuse students, not sure :v

[3man75]

actually it is [-5,5]

So i have to say "divide that X with 25 to get one X and a 1/2 of 25"?

Another question would be: 4x^+9=12x

I move the 12x to the other side by subtracting.

i get 4x^-12x+9=0

What do?

(assuming ^ means to the second)

Also if anyone knows how to land a research job at the federal government or a think tank shoot me a PM. Common core is kicking my ass.

[Descan]

^ means to the power of. ^2 means to the power of 2, ^3 means to the power of 3, etc. "To the power of" means "multiplied together this many times."

And, no, not divide the x by twenty five. You have to figure out which factors, multiplied together, get the x^2 form of the equation/function. Lots of methods, like dividing the x^2 form by one of the factors can get you the other one, if you already know one factor. Or if you have a reasonable guess, you can divide it and see if you get an answer without a remainder. (If there is a remainder, it means the factor you chose was not one of the actual factors.)

Aaah I was never good at explaining things, and especially for slightly more complicated ones, like that 4x equation. v_v

[3man75]

oh um. What if..

X^ - 25=0

5X(-5X)=0  Here i try to have cummualtive multiplication. If i complete the foil thing i get -25 right?

[Descan]

No, you'll just end up with -25x^2, which is -25 times x^2. Not the same thing.

[3man75]

Descan how would you write that mid part?

I'm thinking X(  ) - (  )=0 but i'm not sure. The spaces in the middle of the parenthesis mean nothing btw its just something i made just now.

[Descan]

Which one? I'm a bit confused on what "middle part" you're talkin' about :0

[3man75]

4x^-12x+9=0

Show me the next step from there, if you can please.

[Karlito]

4x²-12x+9 can be factored into the expression (2x-3)(2x-3)=(2x-3)², if that's what you're asking.