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Solving Equations with Fractional Exponents

  • Thread starter Peter G.
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  • #1
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The question is x2/3 - x1/3 - 2 = 0

So the first thing I did was: x2/3 - x1/3 = 2

Then, I put both sides to the power of three, so I got:

x2 - x1 = 8

From there I factorized: x (x - 1) =8

And got the answers: x = 8 or x = 9, the book however, says the correct answers are -1 and 8.

Any help?

Thanks in advance,
Peter G.
 

Answers and Replies

  • #2
33,631
5,288


The question is x2/3 - x1/3 - 2 = 0

So the first thing I did was: x2/3 - x1/3 = 2

Then, I put both sides to the power of three, so I got:

x2 - x1 = 8
Sorry to burst your bubble, but the step above is wrong. When you cube both sides you have to cube the entire side, not just cube individual terms.

The correct approach to this problem is to realize that your initial equation is quadratic in form. Then make the appropriate substitution and solve the resulting quadratic equation.
From there I factorized: x (x - 1) =8

And got the answers: x = 8 or x = 9, the book however, says the correct answers are -1 and 8.

Any help?

Thanks in advance,
Peter G.
 
  • #3
62
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Try u-substitution. Of course it can be solved without doing substitution, but it makes it much easier.
 
  • #4
442
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Sorry, I am a bit confused. What do you mean by cubing the whole side?

I have another similar problem to this one to do next, do you think you could guide me through this one in a bit more detail so I can tackle the next on my own?

Thanks,
Peter G.
 
  • #5
33,631
5,288


You have another mistake, as well.
From there I factorized: x (x - 1) =8

And got the answers: x = 8 or x = 9, the book however, says the correct answers are -1 and 8.
The fact that two numbers multiply to give 8 doesn't really help you very much. If you know that two numbers multiply to 0, then one of them has to be 0.

Your solutions of 8 and 9 don't work. If x = 8, then 8(7) = 8 is not true, so x = 8 is not a solution.

If x = 9, then 9(8) = 8 is also untrue, so x = 9 is also not a solution.
 
  • #6
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u = x1/3

u2 = x2/3

Try that...
 
  • #7
33,631
5,288


Sorry, I am a bit confused. What do you mean by cubing the whole side?

I have another similar problem to this one to do next, do you think you could guide me through this one in a bit more detail so I can tackle the next on my own?

Thanks,
Peter G.
It's probably easier with an example.

x1/2 + 2 = 6

I can square both sides of this equation to get a new equation. The key here is that I have to square (x1/2 + 2), which is different from squaring x1/2 and 2 separately.

After squaring both sides I get

(x1/2 + 2)2 = 62

or x + 4x1/2 + 4 = 36
or x + 4x1/2 = 32

If I rewrite this as x + 4x1/2 - 32 = 0, then I might recognize that this equation is quadratic in form, namely

(x1/2)2 + 4x1/2 - 32 = 0

If I replace x1/2 by u, then I have a true quadratic equation, namely u4 + 4u - 32 = 0.

I can use the quadratic formula or, more simply, just factor the equation above as
(u - 4)(u + 8) = 0

This says that u = 4 or u = -8.

Undoing the substitution I have x1/2 = 4, so x = 16
or x1/2 = -8, (not possible for real numbers x).

As a check, 161/2 + 2 = 4 + 2 = 6, so it checks.

I wouldn't have worked this problem the way I did, since I made more work than was necessary, but I was trying to make a point about squaring both sides. The same applies when you cube both sides, raise both sides to the fourth power, etc.
 
  • #8
442
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Thanks, got what you meant. I will try now.:smile:
 

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