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Algebra problem

C

coder1

Senior member
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?
 
I appreciate the help, but this is still college algebra and I haven't got to Logarithms yet? Is there any other way?
 
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.
 
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.
 
This is only one solution. There are multiple solutions only to square roots. The answer to that is 5.
 
well unless you state the problem, nobody can help you.
Click to expand...

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
 
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.
 
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)
Click to expand...

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.
 
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:
 
Originally posted by: coder1
well unless you state the problem, nobody can help you.
Click to expand...

the acutal problem is x^4-256=0
Click to expand...
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.
 
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
Click to expand...

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

 
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)
 
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)
Click to expand...
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.
 
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)
Click to expand...
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.
Click to expand...

LOL! be nice!

 
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