Week #104: Solving Equation with Real Parameters - March 24, 2014

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SUMMARY

The discussion focuses on solving the equation $\sqrt{x+k}+\sqrt{x+m}+\sqrt{x+n}=\sqrt{x+k+m-n}$, where $k$, $m$, and $n$ are real parameters. Members mente oscura and lfdahl provided correct solutions to this equation. The solutions demonstrate the application of algebraic manipulation and properties of square roots to simplify and solve the equation effectively.

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Solve the equation $\sqrt{x+k}+\sqrt{x+m}+\sqrt{x+n}=\sqrt{x+k+m-n}$ where $k,\,m,\,n$ are real parameters.

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Re: Problem of the week #104 -March 24th, 2014

Congratulations to the following members for their correct solutions::)

1. mente oscura
2. lfdahl

Solution from lfdahl:
\[\sqrt{x+k}+\sqrt{x+m}+\sqrt{x+n}=\sqrt{x+k+m-n} \\\\ \Leftrightarrow (\sqrt{x+k}+\sqrt{x+m})^2=(\sqrt{x+k+m-n}-\sqrt{x+n})^2\\\\ \Leftrightarrow 2x+k+m+2\sqrt{(x+k)(x+m)} = 2x+k+m-n+n-2\sqrt{(x+n)(x-n+k+m)}\\\\ \Leftrightarrow \sqrt{(x+k)(x+m)} = -\sqrt{(x+n)(x-n+k+m)} \\\\ \Rightarrow (x+k)(x+m) = (x+n)(x-n+k+m) \Rightarrow (n-k)(n-m)=0 \\\\ (a). \;\;\;n = k\neq m \Rightarrow x = -n =-k \\\\ (b).\;\;\;n = m \neq k\Rightarrow x = -n=-m \\\\ (c).\;\;\;n=k=m \Rightarrow x = -n=-m=-k\]
 

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