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
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.
You can put this solution on YOUR website! Please help me solve this problem:
:
Why don't you just do this, cube both sides =
:
Gets rid of the denominator in the exponent =
: =
:
Divide both sides by y^3 = -27
y = -27
:
Check solution in calc:
(-27)^(4/3) = 81
and
-3*-27 = 81
:
(