SOLUTION: Please help us solve... (2z+3)^(2/3) + (2z+3)^(1/3) = 6 We've tried cubing each side which gets rather complicated, and we've tried factoring out a (2z+3)^(1/3), but we're no

Question 101751This question is from textbook Precalculus with Limits
: Please help us solve...
(2z+3)^(2/3) + (2z+3)^(1/3) = 6
We've tried cubing each side which gets rather complicated, and we've tried factoring out a (2z+3)^(1/3), but we're not making any progress. Dad is helping, but he's having a tough time remembering complex radical problems! This question is from textbook Precalculus with Limits

Found 2 solutions by bucky, jim_thompson5910:
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
Interesting problem. You can make the problem a little easier to work by making the following
substitution: let
.
When you make that substitution the problem becomes:
.

.
Get this into the standard quadratic form by subtracting 6 from both sides. When you do that
the equation becomes:
.

.
This equation can be solved by factoring. When factored you get:
.

.
This can be solved for A by setting each of the factors equal to zero, which is one technique
for solving a quadratic equation that can be factored. Setting each of the factors equal
to zero results in:
.
which, after subtracting 3 from both sides, gives
.
and in:
.
which, after adding 2 to both sides, results in
.
So there are two answers for A. And at this time we can return to the original definition
of A and substituting into the two answers that we got for A. When
we do we get:
.

.
Cube both sides of this to get:
.

.
Subtract 3 from both sides and you get:
.

.
and solve for z by dividing both sides by 2 to find that:
.

.
You can next use that same process on the second answer that you got for A, which was
A = 2. Substitute for A and you have:
.

.
Cube both sides and you have:
.

.
Subtract 3 from both sides to get:
.

.
Divide both sides by 2 and you get:
.

.
So you have two answers for z ... and
.
Hope this helps you to see how you can get answers for z.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Start with the given equation

Let

So now we get this equation in terms of y

******************************************
Here's an important side note: Take and subtract from both sides to get . This will be useful later
******************************************

Now let's move back to the main problem:

Factor

Break down the root

Factor out

Now divide both sides by

Cube both sides

Expand the left side

Remember, , so replace with

Distribute

Notice the terms and cancel to zero

Combine like terms

Multiply both sides by the LCD y to clear any fractions

Get all terms to one side

Now simply use any technique to solve for y

When you solve for y (I simply used a calculator), you get

or

Now plug in the first answer into

Now solve for z

which is

Now plug in the second answer into

Now solve for z

Check:

Let's check the answer
Start with the given equation

Plug in

Multiply 2 and to get 5

Add 5 and 3 to get 8

Square 8 to get 64

Take the cube root of 64 to get 4 and take the cube root of 8 to get 2

Since the two sides of the equation are equal, this verifies our answer.

Now let's check the answer
Start with the given equation

Plug in

Multiply 2 and to get -30

Add -30 and 3 to get -27

Square -27 to get 729

Take the cube root of 729 to get 9 and take the cube root of -27 to get -3

Since the two sides of the equation are equal, this verifies our answer.

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