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Need some algebra help
- Thread starter rdf8585
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Oh I think I butchered this. I suck at math. It's not done. It might have been written wrong. If I got it wrong don't help me with it I was just bored.rdf8585 said:
Edit: Yeah it was written wrong..
I think you made that way more complicated.....mine was complicated, but not that much.
PS You don't want my answer...Mine came out to 5 = 1.
PS You don't want my answer...Mine came out to 5 = 1.
the other one looks like
http://img212.imageshack.us/my.php?image=untitled2ov8.png
if theres any confusion, its the 5th root of...
solve for z
http://img212.imageshack.us/my.php?image=untitled2ov8.png
if theres any confusion, its the 5th root of...
solve for z
Use the solve function in a bloody calculator. Single variable algebra problems should not warrant much of your time, and they won't later on in life. You should still know how to do it, but don't idle on it for too long.
How old are you? Grade?
How old are you? Grade?
x+(sqrt33-2x^2) = 3
x + sqrt(33-2x^2) = 3 // if you meant this, then
x = -2
12+ the 5th root of (z^3+63) = 11
12+ (z^3+63)^(1/5) = 11
x = null //no results
Sorry if I made a mistake here somewhere, or understood the questions wrong... The calculations were solved with the "Solve[]" function of Mathematica 5.1. But right now I am in no mental condition do double-check these by hand.
x + sqrt(33-2x^2) = 3 // if you meant this, then
x = -2
12+ the 5th root of (z^3+63) = 11
12+ (z^3+63)^(1/5) = 11
x = null //no results
Sorry if I made a mistake here somewhere, or understood the questions wrong... The calculations were solved with the "Solve[]" function of Mathematica 5.1. But right now I am in no mental condition do double-check these by hand.
Okay...let's start with the first one. Rather than just give you the answer, how about we figure out how it should be done?
1. You want to get rid of that nasty square root. But in order to do that you have to get rid of that darn extra x on the left hand side. That's pretty easy.
2. Now let's square both sides of the equation. The left side is easy...the square root just goes poof. The right side is a little more challenging, but if you remember your FOIL rule, you'll be in good shape.
3. Now you'll have a bunch of x^2, x, and constant terms. Move things around to combine like terms. Move everything to one side and you've got a quadratic equation to solve.
1. You want to get rid of that nasty square root. But in order to do that you have to get rid of that darn extra x on the left hand side. That's pretty easy.
2. Now let's square both sides of the equation. The left side is easy...the square root just goes poof. The right side is a little more challenging, but if you remember your FOIL rule, you'll be in good shape.
3. Now you'll have a bunch of x^2, x, and constant terms. Move things around to combine like terms. Move everything to one side and you've got a quadratic equation to solve.
ChrisBrightwell said:
Whoops. Well, z is null, too (Actually, I think I just made a mistake tying the output from the MacBook Pro to the iMac. I don't have Mathematica installed on this one).
Here are my answers, but I'm really rusty, so they need to be double checked.
1) -2
2) 4i
They are probably wrong.
1) -2
2) 4i
They are probably wrong.
Here's the solution for the second problem. Hope you can see it. Algebra is so fun.
12+5th root z^3+63=11
5th root z^3+63=-1
Set both sides to the power of 5 (5th root should cancel out with the exponent)
z^3+63=-1
z^3=-64
z=-4
EDIT: Whoops...someone already got to it.
12+5th root z^3+63=11
5th root z^3+63=-1
Set both sides to the power of 5 (5th root should cancel out with the exponent)
z^3+63=-1
z^3=-64
z=-4
EDIT: Whoops...someone already got to it.
EricNau said:I could have sworn you could not root a negative number (making it imaginary).
What did I miss?
z^3=-64
The exponent is odd, which allows for negatives. Put it in a better way,
z=-4
z^3=-4 x -4 x -4=-64
Taking a even root from a negative number will produce a imaginary number.
OK... That makes sense, but why doesn't the calculator give that answer? It says "Not a Number"rontheancient said:
EricNau said:OK... That makes sense, but why doesn't the calculator give that answer? It says "Not a Number"
What type of calculator are you using? Alternatively, instead of using a cube root button you can set -64 to the 1/3 power, but just make sure 1/3 is in parentheses.
EricNau said:
Oh, thats a bit ugly. I've been trying to calculate the cube root of -64 but I get the same answer as you. It works fine on my graphing calc.