solve irrational eqn.: (x-9)^(1/3) - (x+26)^(1/3) + 5 = 0

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Luke53
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solve irrational eqn.: (x-9)^(1/3) - (x+26)^(1/3) + 5 = 0

Postby Luke53 » Tue Mar 13, 2012 2:17 pm

(x-9)^(1/3) - (x+26)^(1/3) + 5 = 0
How to solve this one?
Thanks.
Luke.

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stapel_eliz
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Postby stapel_eliz » Wed Mar 14, 2012 2:03 pm

I was going to suggest cubing both sides, but that doesn't seem to help much. Are you sure that you have copied the exercise correctly? I haven't gone very far in attempting to find a solution method, and I'd like to verify the exercise statement before proceeding much further.

Thank you! :wink:

Luke53
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Re: solve irrational eqn.: (x-9)^(1/3) - (x+26)^(1/3) + 5 =

Postby Luke53 » Wed Mar 14, 2012 6:46 pm

Hi, I checked the equation and there are no errors, when I solve this eqn. with my graphic calculator (TI89) it gives me two roots 1 and -18, but sadly this thing only gives me the results and not how to get to them.
According to my textbook it should be done like you mentioned by cubing both members. My textbook states that: A^3 - B^3 = 0 is the same as (A - B) * (A^2+AB+B^2) = 0 witch falls apart in A - B = 0 and
A^2 + AB + B^2 = 0 the eqn. A^2 + A B + B^2 has no roots except those that make A and B zero at the same time; but also statisfy A = B. The solutions A^3 = B^3 are also solutions of A = B and vice versa.
The eqn. A^(1/3)+B^(1/3) = c^(1/3) becomes rational if we do the following: Cube both members to get: A + B +3*(A B)^(1/3) * A^(1/3) + B^(1/3) = C or 3(A B C)^(1/3) = C - A - B cubing the last eqn again gives: 27 A B C = (C - A - B)^3.
Very nice so far, but what are A, B and C in the eqn to solve? I suppose A= (x-9)^(1/3); B = (x+26)^(1/3) and C = 5. (but that does not mean I am able solve this yet).
Sorry for errors, I had to translate this text.

Regards.

Luke.

FWT
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Re: solve irrational eqn.: (x-9)^(1/3) - (x+26)^(1/3) + 5 =

Postby FWT » Wed Mar 14, 2012 11:24 pm

Luke53 wrote:According to my textbook it should be done like you mentioned by cubing both members.

I am not aware of any legitimate solution method which would involve cubing two of the terms. One can only cube both entire sides.

I do not see the sense of the rest of what the text appears to say. Moving the 5 to the right-hand side and cubing both sides gives:





I do not see how your text then derives an equation with only one cube root on the left-hand side.


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