Need some algebra help

Discussion in 'Community Discussion' started by rdf8585, Jul 22, 2006.

1. rdf8585 macrumors regular

Joined:
Feb 15, 2006
#1
Solve for x
x+(sqrt33-2x^2) = 3

Solve for x
12+ the 5th root of z^3+63 = 11

Thanks to anyone who can help. I'd really appreciate your time here.

2. paleck macrumors 6502a

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with the Tequila!
#2
Should that be x+sqrt(33-(2x)^2) = 3?

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4. Coolnat2004 macrumors 6502

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Jan 12, 2005
#4
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.

Edit: Yeah it was written wrong..

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6. paleck macrumors 6502a

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#6
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.

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Feb 15, 2006
#7
8. cleanup macrumors 68030

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Toronto
#8
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.

9. rdf8585 thread starter macrumors regular

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Feb 15, 2006
#9
I'm in College Algebra II.... I failed this stuff in the spring, re-taking it now. I don't have a graphing calc, if thats what you mean.

Again, I could really use some help on these 2 problems.

10. Kaioshin macrumors member

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otanj, Slovenia
#10
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

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.

Staff Member

Joined:
Jan 20, 2005
#11
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.

12. rdf8585 thread starter macrumors regular

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Feb 15, 2006
#12
Yeah, the first one makes more sense now. FOIL and solving most qadratics don't bother me. But the second question, the one with a 5th root, is a doozy.

13. ChrisBrightwell macrumors 68020

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Huntsville, AL
#13
Of course x is null. He's trying to solve for z.

14. Kaioshin macrumors member

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Jul 17, 2006
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otanj, Slovenia
#14
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).

15. EricNau Moderator emeritus

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Apr 27, 2005
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San Francisco, CA
#15
To be fair, he did say "solve for x" - so you answered his question!

16. hoyboy9 macrumors member

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Oct 1, 2004
#16
Engineer to the rescue.

5th(z^3+63) = -1

Take each side to the 5th power:

z^3+63 = -1
z^3 = -64

Take the cubed root of both sides:

z = -4

Enjoy.

17. EricNau Moderator emeritus

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Apr 27, 2005
Location:
San Francisco, CA
#17
Here are my answers, but I'm really rusty, so they need to be double checked.

1) -2

2) 4i

They are probably wrong.

18. rontheancient macrumors regular

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Nov 22, 2005
Location:
USA
#18
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.

19. EricNau Moderator emeritus

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#19
I could have sworn you could not root a negative number (making it imaginary).

What did I miss?

20. rdf8585 thread starter macrumors regular

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Feb 15, 2006
#20
Thanks again to everyone here. Appreciate it.

21. rontheancient macrumors regular

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USA
#21
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.

22. EricNau Moderator emeritus

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San Francisco, CA
#22
OK... That makes sense, but why doesn't the calculator give that answer? It says "Not a Number"

23. rontheancient macrumors regular

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Nov 22, 2005
Location:
USA
#23

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.

24. EricNau Moderator emeritus

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Apr 27, 2005
Location:
San Francisco, CA
#24
I was using the Apple calculator (the one that comes with every Mac).

25. rontheancient macrumors regular

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Nov 22, 2005
Location:
USA
#25

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.