SOLUTION: Please help me solve this problem: y^4/3 = -3y This is what I have done: y^4/3 + 3y =0 1. The exp of y is understood to be 1...so does this mean that I need to find the c

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Please help me solve this problem: y^4/3 = -3y This is what I have done: y^4/3 + 3y =0 1. The exp of y is understood to be 1...so does this mean that I need to find the c      Log On


   

Question 96205This question is from textbook Fundamentals of Alg and Trig
: Please help me solve this problem:
y^4/3 = -3y
This is what I have done:
y^4/3 + 3y =0
1. The exp of y is understood to be 1...so does this mean that I need to find the commoon denominator of the exp so that I could add? Should the exp of 3y then be 1/3?
y^4/3 + 3y^1/3= 0
2. Then in the book it says to factor out maybe 1/3?
This is where I get confused, because the problem in the book that is simmilar looks like this:
x^3/2= x^1/2
x^3/2- x^1/2= 0
x^1/2(x-1)) =0 How are they factoring out x^1/2? Is x^1/2(x-1) = x^3/2 -x^1/2
3. So once they get x^1/2= 0 and x-1=0
so the answer is x=0 and x=1 This question is from textbook Fundamentals of Alg and Trig

Found 4 solutions by stanbon, ankor@dixie-net.com, MathTherapy, greenestamps:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
y^4/3 = -3y
-------------------
Comment: It's always the lowest power that you factor out.
Example: If you have x^3+x^2 you factor out x^2 to get x^2(x+1)
If you have x^-5 + x^-2 you factor out x^-5 to get x^-5(1+x^3)
-----------------
Your Problem
Rearrange the equation:
y^(4/3)+3y=0
The lowest power is y so you factor it out:
y(y^(1/3) + 3) = 0
y = 0 or y^(1/3)+3 = 0
y=0 or y= (-3)^3
y=0 or y = -27
======================
Cheers,
Stan H.

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Please help me solve this problem:
y%5E%284%2F3%29+=+-3y
:
Why don't you just do this, cube both sides
%28y%5E%284%2F3%29%29%5E3 = %28-3y%29%5E3
:
Gets rid of the denominator in the exponent
y%5E4= -3%5E3%2Ay%5E3
:
y%5E4= -27%2Ay%5E3
:
Divide both sides by y^3
%28y%5E4%29%2F%28y%5E3%29 = -27
y = -27
:
Check solution in calc:
(-27)^(4/3) = 81
and
-3*-27 = 81
:

Answer by MathTherapy(10660) About Me  (Show Source):
You can put this solution on YOUR website!
Please help me solve this problem:

y^4/3 = -3y

This is what I have done:
y^4/3 + 3y =0

1. The exp of y is understood to be 1...so does this mean that I need to find the commoon
denominator of the exp so that I could add?  Should the exp of 3y then be 1/3?
y^4/3 + 3y^1/3= 0

2. Then in the book it says to factor out maybe 1/3?

This is where I get confused, because the problem in the book that is simmilar looks like this:
x^3/2= x^1/2
x^3/2- x^1/2= 0
x^1/2(x-1)) =0   How are they factoring out x^1/2?  Is x^1/2(x-1) = x^3/2 -x^1/2

3. So once they get x^1/2= 0   and x-1=0

so the answer is x=0 and x=1 

I think you're CONFUSING yourself, and probably mixing up this given problem with the one in your book.
Anyway, I hope you can follow what I've presented below. 

            matrix%282%2C1%2C+%22+%22%2C+y%5E%284%2F3%29+=+-+3y%29
        matrix%282%2C1%2C+%22+%22%2C+y%5E%284%2F3%29%29+%2B+3y+=+0 ---- Adding 3y%5E2 to both sides
     matrix%282%2C1%2C+%22+%22%2C+y%28y%5E%281%2F3%29+%2B+3%29+=+0%29 ---- Factoring out y
         matrix%282%2C1%2C+%22+%22%2C+y%5E%281%2F3%29%29+%2B+3+=+0    OR     y = 0  
           matrix%282%2C1%2C+%22+%22%2C+y%5E%281%2F3%29%29+=+-+3
       matrix%282%2C1%2C+%22+%22%2C+y%5E%28%281%2F3%29%5E3%29%29+=+%28-+3%29%5E3 --- Cubing each side
              highlight%28y%29+=+%28-+3%29%5E3+=+highlight%28-+27%29%29

Answer by greenestamps(13276) About Me  (Show Source):
You can put this solution on YOUR website!


...

This is what I have done:
y^4/3 + 3y =0
1. The exp of y is understood to be 1...so does this mean that I need to find the commoon denominator of the exp so that I could add? Should the exp of 3y then be 1/3?
y^4/3 + 3y^1/3= 0


No.

As the other tutor says, you seem to be trying to make this problem look like the example in your book, but it is very different.

In the expression "3y" in this problem, the understood exponent "1" is only the exponent of "y" -- it is not the exponent of "3y".

Leave the equation as

y^4/3 + 3y =0

Then, as the other tutor shows, factor out the common factor "y" on the left and solve the problem from there.