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October 24th, 2015, 08:27 PM   #1
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when is isolating a variable impossible?

Say I have an algebraic equation in one or more variables. Under what circumstances is it impossible to isolate all instances of one of the variables on one side of the equation? Are there any theorems that say when it isn't possible or is this simply too broad a question?
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October 25th, 2015, 04:44 AM   #2
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Largely, that depends upon exactly what you mean. For example, starting with the equation $\displaystyle xe^x= a$, there are no "elementary" methods for isolating the "x". But the "Lambert W function", W(x), is specifically defined as the inverse function to $\displaystyle f(x)= xe^x$ so that $\displaystyle x= W(a)$.

More generally, if we are given some formula in "x" and want to be able write "x= " some single value, then that formula must be "one to one". For something as simple as $\displaystyle y= x^2$ we cannot "isolate" x since we would have to choose between $\displaystyle x= \sqrt{y}$ and $\displaystyle x= -\sqrt{y}$.

Or if we have something like (a- b)x= c, we can write $\displaystyle x= \frac{c}{a- b}$ as long as a is not equal to b. If a= b, the equation becomes 0x= 0= c which no longer has an "x" in it!
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Last edited by Country Boy; October 25th, 2015 at 04:47 AM.
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October 25th, 2015, 05:32 AM   #3
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Thanks Country Boy. I guess I should have been more specific about what I meant by "isolating x". For the y = x^2 example, I would consider x to be isolated even though there is more than one solution.

The reason for my question is that an equation like $\displaystyle \frac{\sqrt{a+x}}{x}=\frac{\sqrt{b}}{x+h}$ has solutions that seem like they would be very difficult to find manually.

They are shown here: solve for x sqrt(a+x)/x = sqrt(b)/(x+h)–Wolfram|Alpha

If I was asked to find a solution to the equation above, how would I know if it is even possible? And how would one even go about solving this equation for x?

Last edited by skipjack; October 25th, 2015 at 07:06 PM.
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October 25th, 2015, 07:03 PM   #4
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Square both sides of the equation.
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October 26th, 2015, 05:41 PM   #5
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Quote:
Originally Posted by skipjack View Post
Square both sides of the equation.
If I did it right I get:
$\displaystyle bx^2 = ax^2 + ahx + ah^2 + x^3 + 2hx^2 + h^2x$
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