BF2S Forums How good are you at algebra?

stef10 wrote:

Ganko_06 wrote:

Try some Calculus

y = (x+1)^(x+1)^(x+1)  Find y'

WTF! I have never seen this icon before  ^

"to the power of ...." symbol.
Indices crap.

stef10 wrote:

Okay I want you guys to be honest. When at school did you learn this.

8th grade

One root of:

z^4 - 10z^3 + 42z^2 - 82z + 65 = 0

Is:

3 + 2j

---

Solve the equation, hence, find the other 3 roots.

(3 + 2j) is a root, thus it's complex conjugate (3 - 2j) is also.

hence, (z - 3 - 2j) and (z - 3 + 2j) are factors of z^4 - 10z^3 + 42z^2 - 82z + 65 = 0

thus, (z^2 - 6z + 13) is a factor (multiplying the previos 2 together)

working out  (z^4 - 10z^3 + 42z^2 - 82z + 65) / (z^2 - 6z + 13)

gives (z^2 - 4z + 5)

solving this with the quadratic formula gives:

(z - 2 - j) and (z - 2 + j)

or: z = (2-j), (2+j)

this gives you the final factorisation:
(z - 3 - 2j)(z - 3 + 2j)(z - 2 - j)(z - 2 + j)

or otherwise, the 4 roots are:

z = (3 + 2j), (3 - 2j), (2 - j), (2 + j)

Now what do I win

Edit: formatted roots correctly

Last edited by munchagepeople (2006-10-21 14:04:54)

Took me a little while longer, but this is how it's done:

Try some Calculus

y = (x+1)^(x+1)^(x+1)  Find y'

y = (x+1)^(x+1)^(x+1) = (x+1)^g(x) where g(x) = (x+1)^(x+1)

then, ln(y) = ln(x+1)^g(x)

-> ln(y) = g(x)*ln(x+1)

now, using implicit differentiation and the product rule:

(1/y)f'(x) = g'(x)*ln(x+1) + g(x)*(1/(x+1))

rearranging and substitution gives:
f(x) = f(x) * {g'(x)*ln(x+1) + ((x+1)^(x+1)/(x+1))}

Now to find g'(x):

g(x) = (x+1)^(x+1)

so:
ln(g(x)) = (x+1)*ln(x+1)

and using product rule and implicit differentiation again:
(1/g(x))*g'(x) = (x+1)/(x+1) + ln(x+1)

-> g'(x) = g(x) + g(x)*ln(x+1)
or: g'(x) = (x+1)^(x+1) + (x+1)^(x+1)*ln(x+1)

substituting this back into the equation for f'(x):

f'(x) = f(x) * {(x+1)^(x+1)[1 + ln(x+1)]*ln(x+1) + (x+1)^(x+1)/(x+1)}

this gives a final result of:

f'(x) = (x+1)^(x+1)^(x+1) * {(x+1)^(x+1)[1 + ln(x+1)]*ln(x+1) + (x+1)^(x+1)/(x+1)}

*phew*

I don't know where you got that from, but that question was evil

Any questions about the method, or any steps just ask

Edit: sorted out a closing brace I missed...

Last edited by munchagepeople (2006-10-21 17:10:59)

haha I aint helping you do your homework for free

stef10 wrote:

Well I am pretty bored right now so i decided to go on a website and practice some algebra for next week.
Let us see if you can figure these out. A reminder. These should be pretty easy.

Just take them you want. I will tell the answers later or you could just pm the answers. karma if you do well.

1: 2(x - 3) = x + 4

2: 5x - (12 + 3x) = 2

3: 15 - (6 - 8x) = 25

4: 2(x - 3) + 5(x + 1) = 34

5: (6x - 8) - (x + 4) = 6 - x

6: (5 - x) - (10 - 2x) = 6 - x - 5

7: 5(x - 2) + 2(3x + 2) = 27

8: 3(1 - 3x) - (x + 5) = 20

9: 31 + 2(3x - 4) = 5(2x + 3)

10: 12( 2x - 3) - 6(x - 8) = 120

1. x = 10
2. x = 7
3. x = 2
4. x = 5
5. x = 3
6. x = 3
7. x = 3
8. x = -2.2
9. x = 2
10. x = 6

[edit] why did i do this?
when anwsers have been posted a trillion times

Last edited by Gamematt (2006-10-21 17:28:45)

liquidat0r wrote:

pfft. Go do some real maths:

Complex Numbers

One root of:

z^4 - 10z^3 + 42z^2 - 82z + 65 = 0

Is:

3 + 2j

---

Solve the equation, hence, find the other 3 roots.

The roots of the equation are:

3+2i
3-2i
2+i
2-i

(we use i to denote imaginary number over here )

Last edited by Vub (2006-10-21 22:56:23)

Here's an easy one:

Integrate 1/(xlnx) or one on x times lnx.

Ganko_06 wrote:

Vub wrote:

Here's an easy one:

Integrate 1/(xlnx) or one on x times lnx.

Integrated or differentiate?

Integrate

Ah, I'm tempted to post my 3D calculus problems and have some of you help me solve them. Interested? PM me.

pffffft I learnt this at 7th grade.

Stags wrote:

Vub wrote:

Here's an easy one:

Integrate 1/(xlnx) or one on x times lnx.

ln(ln(x))+c 

Lovely, Calculus, grew bored off math in HS so that's probably why I'm a Business Major.


I could be very annoying, y' of e^e^e^e^e^e^e^e^e^e^x  :p  Loved this problem easy points on a test.



Then of course we could do this.  Find the Volume of the resulting object.  Formed from revolving the area between  y = (x+3)^(1/2) and y=x^(x-1) around y=4.  I haven't worked out the math to it (yet) but it shouldn't kill you math buffs out there.

EDIT: Weird Error.  Anyways fixed it.

e^x times e^e^x times e^e^e^x etc. until you get to e^e^e^e^e^e^e^e^e^e^x