Algebra problem

[coder1]

Ok I know how to find the real solutions but what about all solutions. I can't seem to figure out the proper steps

exmaple:

x^4=125

anyone there that can point me in the right direction?

 

[GoodToGo]

Logarithm on both sides.

 

[coder1]

I appreciate the help, but this is still college algebra and I haven't got to Logarithms yet? Is there any other way?

 

[Heisenberg]

What do you mean by all solutions? Both imaginary and real solutions?

 

[coder1]

correct

 

[Heisenberg]

Take both sides to the 1/3 power. 1/3 is the same as 2/6 so you can have 5i as a solution also. I think - I'm on my way out the door so that may not be completely right.

 

[jagr10]

What the hell is there to find? Can't you just put each side to the power of 1/3? It's all real to me. No need to imagine things.

 

[VIAN]

This is only one solution. There are multiple solutions only to square roots. The answer to that is 5.

 

[coder1]

the answers are +4, -4, +4i, -4i I am trying to understand how to get to 4i or sqr(16)*sqr(-1)

 

[shuan24]

well unless you state the problem, nobody can help you.

 

[coder1]

well unless you state the problem, nobody can help you.

the acutal problem is x^4-256=0

I originaly posted x^3-256=0, it is to the 4th power insted , sorry for confusing the hell out of everyone

 

[Howard]

Originally posted by: coder1
the answers are +4, -4, +4i, -4i I am trying to understand how to get to 4i or sqr(16)*sqr(-1)

??? x = +/- 5 in your example!

Oh, new question.

 

[edro]

x=5? hell if I know

 

[Howard]

x^4 - 256 = 0
x^4 = 256
log x^4 = log 256
4 log x = log 256
log x = log 256 / 4
x = 10^(log 256/4)

That's one solution anyway.

 

[eLiu]

Originally posted by: coder1
the answers are +4, -4, +4i, -4i I am trying to understand how to get to 4i or sqr(16)*sqr(-1)

Huh?! But 4^3 is 64. The problem asks for x^3=125, so x=5. There is no imaginary part because i^3 = -i and (-i)^3 = i. Thus, if you cube i, your answer will have an i in it...which 125 clearly does not. The answer here is just x=5. This is a repeated root (in that it "occurs" 3 times).

Edit: Blarr...it said x^3=125, i swear...but now it's x^4=125. Man am I seeing things?

But for this problem...the fourth root of 125 is obviously one answer. The negative of that is another. Now for the imaginary part...let's note that i^4 = 1. So, the fourth root of 125 * i is also an answer...and since imaginary roots always come in conjugate-pairs, the negative the previous answer is also a root.

 

[coder1]

thanks for everyones help, sorry about putting down the incorrect exponent. it is 4 not 3.

 

[shuan24]

Dude, I'll give you a hint:

x^4 - 256 =
(x^2 - 16)*(x^2 +16) =
(x-4)*(x+4)*(x-4)*(x+4) =
(x-4)^2 * (x+4)^2 = 0

Now solve for all values of x.

:beer:

 

[coder1]

now it's making sense.

 

[dullard]

Originally posted by: coder1

well unless you state the problem, nobody can help you.

the acutal problem is x^4-256=0

Starting problem:
x^4 = 256

Take the square root of each side:
x^2 = +- 16

Notice that both +16 and -16 work. Why? Since (+16)*(+16) = 256 and since (-16)*(-16) = 256. Now there are two separate problems:
x^2 = 16 and x^2 = -16

Lets work on the first of those two. Take the square root of both sides:
x = (16)^0.5

Thus:
x = +-4

Thus either x = 4 or x = -4 will work. Check to be sure (4)*(4)*(4)*(4) = 256 and (-4)*(-4)*(-4)*(-4) = 256. Yep got two of the answers. We know there are two more answers since the original equation was a power of x to the 4 (interesting rule, did you know that)? Lets look at the second equation we ignored above:
x^2 = -16

Take the square roots of both sides:
x = +-(-16)^0.5

Now we are dealing with imaginary numbers. Lets pull out the negative sign:
x = +- (16)^0.5*(-1)^0.5

Thus
x = +- 4 * (-1)^0.5

Now the imaginary number i is equal to (-1)^0.5. Thus
x = +4i and x = -4i.

Do the check like I did above, and you'll see those also are correct.

 

[shuan24]

Originally posted by: coder1
now it's making sense. Ok here is the last one, I can now do the first one but this one is got me stuck

x^3=125

wtf? people's been posting the solution to that.

 

[shuan24]

nice approach dullard, I just did it in a different way.

(but by using the quadratic equation it is more versatile than having a perfect root)

 

[jagr10]

Originally posted by: shuan24

Originally posted by: coder1
now it's making sense. Ok here is the last one, I can now do the first one but this one is got me stuck

x^3=125

wtf? people's been posting the solution to that.

lol.

 

[coder1]

Thanks for everyones help, i got it now, sorry for all the mistypes

 

[dullard]

Originally posted by: shuan24
nice approach dullard, I just did it in a different way.

(but by using the quadratic equation it is more versatile than having a perfect root)

Trying to teach someone in college algebra how to factor is like trying to teach a rock to swim. Thus I use the brute force method in situations like this. Of course my method will be difficult to show coder1 that x^3 has three solutions that all happen to be 5. So we must take your approach on that one. Good luck on the swim lessons.

 

[shuan24]

Originally posted by: dullard

Originally posted by: shuan24
nice approach dullard, I just did it in a different way.

(but by using the quadratic equation it is more versatile than having a perfect root)

Trying to teach someone in college algebra how to factor is like trying to teach a rock to swim. Thus I use the brute force method in situations like this. Of course my method will be difficult to show coder1 that x^3 has three solutions that all happen to be 5. So we must take your approach on that one. Good luck on the swim lessons.

LOL! be nice!