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Consider the equation:

$$2\sqrt {2x}+\sqrt {2x+3}=\sqrt {3x+2}+\sqrt {6x+20}.$$

Find a trick ( if exists ) which allows to solve it elegantly i.e. with avoiding the systematic squaring.

(The systematic squaring inevitably leads to a fourth-degree equation:

$$ \begin{align} 0 &= 207x^4-12564x^3+27738x^2+231084x-40401\\[6pt] &=9\left( 23x^2-1258x-4489\right) \left( x^2-6x+1\right)\;, \end{align} $$ so the answer is $$x=\dfrac {629+\sqrt {498888}} {23}.$$

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Why do you think there is a trick? Also, why just the one answer? $x^2 - 6x+1$ has two positive roots, $3 \pm \sqrt 8.$ – Will Jagy Jul 22 '12 at 22:44
@Will You can check with mathematica that there exist unique solution here – Norbert Jul 22 '12 at 22:51
@Will: The squaring can introduce spurious solutions, since the squared equation is satisfied even if the unsquared one had the wrong signs. I get the same quartic when I do the squaring with Wolfram|Alpha. – joriki Jul 22 '12 at 23:01
Alpha… shows only the one real solution – Ross Millikan Jul 22 '12 at 23:39
You want an elegant way to get to $\sqrt{498888}$. I can't imagine such a thing exists. – Gerry Myerson Jul 22 '12 at 23:53
  1. This equasion is defined for all $x>0$.

  2. Function on the left hand side is strictly increasing (i.e always increasing).

  3. Function on the right hand side is strictly increasing (i.e always increasing) too.

We can draw a simple graph on large scale, but with low prescision, to find out where too look for this root. We'll see on the graph that there is 1 intersection. Beacuse both functions strictly increasing there will be no other roots for $x>0$.

You can use this or any other tool to draw two graphs for left and right hand side functions.

From that graph we'll see that $x \in (58.02 , 58.08)$. You can get the decimal answer depending on the prescision required by incrasing prescision of interval ends.

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@RossMillikan, I agree, I edited my answer. – Stoleg May 24 '13 at 13:55

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