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

Algebra ->  Radicals -> 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      Log On


   

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) About Me  (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 A+=+%282z+%2B+3%29%5E%281%2F3%29
.
When you make that substitution the problem becomes:
.
A%5E2+%2B+A+=+6
.
Get this into the standard quadratic form by subtracting 6 from both sides. When you do that
the equation becomes:
.
A%5E2+%2B+A+-+6+=+0
.
This equation can be solved by factoring. When factored you get:
.
%28A+%2B+3%29%2A%28A+-+2%29=+0
.
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:
.
A+%2B+3+=+0 which, after subtracting 3 from both sides, gives A+=+-3
.
and in:
.
A+-+2+=+0 which, after adding 2 to both sides, results in A+=+2
.
So there are two answers for A. And at this time we can return to the original definition
of A and substituting %282z+%2B+3%29%5E%281%2F3%29 into the two answers that we got for A. When
we do we get:
.
%282z+%2B+3%29%5E%281%2F3%29+=+-3
.
Cube both sides of this to get:
.
2z+%2B+3+=+-27
.
Subtract 3 from both sides and you get:
.
2z+=+-30
.
and solve for z by dividing both sides by 2 to find that:
.
z+=+-15
.
You can next use that same process on the second answer that you got for A, which was
A = 2. Substitute %282z+%2B+3%29%5E%281%2F3%29 for A and you have:
.
%282z+%2B+3%29%5E%281%2F3%29+=+2
.
Cube both sides and you have:
.
2z+%2B+3+=+8
.
Subtract 3 from both sides to get:
.
2z+=+5
.
Divide both sides by 2 and you get:
.
z+=+5%2F2
.
So you have two answers for z ... z+=+-15 and z+=+5%2F2
.
Hope this helps you to see how you can get answers for z.

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
root%283%2C%282z%2B3%29%5E2%29%2Broot%283%2C%282z%2B3%29%29=6 Start with the given equation

Let y=2z%2B3


root%283%2Cy%5E2%29%2Broot%283%2Cy%29=6 So now we get this equation in terms of y


******************************************
Here's an important side note: Take root%283%2Cy%5E2%29%2Broot%283%2Cy%29=6 and subtract root%283%2Cy%29 from both sides to get root%283%2Cy%5E2%29=6-root%283%2Cy%29. This will be useful later
******************************************

Now let's move back to the main problem:

root%283%2Cy%2Ay%29%2Broot%283%2Cy%29=6 Factor y%5E2


root%283%2Cy%29%2Aroot%283%2Cy%29%2Broot%283%2Cy%29=6 Break down the root


root%283%2Cy%29%28root%283%2Cy%29%2B1%29=6 Factor out root%283%2Cy%29


%28root%283%2Cy%29%2B1%29=6%2Froot%283%2Cy%29 Now divide both sides by root%283%2Cy%29


%28root%283%2Cy%29%2B1%29%5E3=%286%2Froot%283%2Cy%29%29%5E3 Cube both sides


y%2B3root%283%2Cy%29%2B3root%283%2Cy%5E2%29%2B1=216%2Fy Expand the left side


Remember, root%283%2Cy%5E2%29=6-root%283%2Cy%29, so replace root%283%2Cy%5E2%29 with 6-root%283%2Cy%29


y%2B3root%283%2Cy%29%2B3%2Ahighlight%28%286-root%283%2Cy%29%29%29%2B1=216%2Fy


y%2B3root%283%2Cy%29%2B18-3root%283%2Cy%29%2B1=216%2Fy Distribute


y%2B18%2Bcross%283root%283%2Cy%29-3root%283%2Cy%29%29%2B1=216%2Fy Notice the terms 3root%283%2Cy%29 and -3root%283%2Cy%29 cancel to zero


y%2B18%2B1=216%2Fy


y%2B19=216%2Fy Combine like terms



y%5E2%2B19y%5E2=216 Multiply both sides by the LCD y to clear any fractions



y%5E2%2B19y%5E2-216=0 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

y=8 or y=-27


Now plug in the first answer y=8 into y=2z%2B3


8=2z%2B3


Now solve for z

z=5%2F2 which is z=2.5


Now plug in the second answer y=-27 into y=2z%2B3


-27=2z%2B3


Now solve for z

z=-15




Check:


Let's check the answer z=5%2F2
root%283%2C%282z%2B3%29%5E2%29%2Broot%283%2C%282z%2B3%29%29=6 Start with the given equation



root%283%2C%282%285%2F2%29%2B3%29%5E2%29%2Broot%283%2C%282%285%2F2%29%2B3%29%29=6 Plug in z=5%2F2



root%283%2C%285%2B3%29%5E2%29%2Broot%283%2C%285%2B3%29%29=6 Multiply 2 and 5%2F2 to get 5



root%283%2C8%5E2%29%2Broot%283%2C8%29=6 Add 5 and 3 to get 8


root%283%2C64%29%2Broot%283%2C8%29=6 Square 8 to get 64



4%2B2=6 Take the cube root of 64 to get 4 and take the cube root of 8 to get 2


6=6 Since the two sides of the equation are equal, this verifies our answer.





Now let's check the answer z=-15
root%283%2C%282z%2B3%29%5E2%29%2Broot%283%2C%282z%2B3%29%29=6 Start with the given equation



root%283%2C%282%28-15%29%2B3%29%5E2%29%2Broot%283%2C%282%28-15%29%2B3%29%29=6 Plug in z=-15



root%283%2C%28-30%2B3%29%5E2%29%2Broot%283%2C%28-30%2B3%29%29=6 Multiply 2 and -15 to get -30



root%283%2C%28-27%29%5E2%29%2Broot%283%2C-27%29=6 Add -30 and 3 to get -27


root%283%2C729%29%2Broot%283%2C-27%29=6 Square -27 to get 729



9-3=6 Take the cube root of 729 to get 9 and take the cube root of -27 to get -3


6=6 Since the two sides of the equation are equal, this verifies our answer.