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#### sweatingbear

##### Member

- May 3, 2013

- 91

\(\displaystyle \sqrt{x+13} - \sqrt{7-x} = 2 \, .\)

If \(\displaystyle x \in \mathbb{R}\) then we must require that

\(\displaystyle \begin{cases}

x+13 \geqslant 0 & \iff & x \geqslant -13 \\

7-x \geqslant 0 & \iff & x \leqslant 7

\end{cases}\)

Moreover, if we rewrite equation as \(\displaystyle \sqrt{x+13} = 2 + \sqrt{7-x}\), the right-hand side must in fact be greater than or equal to zero (since the output of \(\displaystyle \sqrt{x+13}\) is greater than or equal to zero). Thus

\(\displaystyle 2 + \sqrt{7-x} \geqslant 0 \iff \sqrt{7-x} \geqslant -2 \, .\)

but \(\displaystyle \sqrt{7-x} \geqslant -2 \ , \ \forall x \in \mathbb{R}\) so no further restriction on \(\displaystyle x\) could be derived.

We could rewrite the equation as \(\displaystyle \sqrt{7-x} = \sqrt{x+13} - 2\) and apply a similar reasoning about the output of the square root function. We must require \(\displaystyle \sqrt{x+13} - 2 \geqslant 0\).

\(\displaystyle \sqrt{x+13} -2 \geqslant 0 \iff \sqrt{x+13} \geqslant 2 \iff x+13 \geqslant 4 \iff x \geqslant -9 \, .\)

So, the restrictions we have on \(\displaystyle x\) thus far are

\(\displaystyle (x \geqslant -13) \wedge (x \leqslant 7) \wedge (x\geqslant -9) \implies -9 \leqslant x \leqslant 7 \, .\)

Solving the aforementioned equation by squaring a few times and using the quadratic formula yields the roots \(\displaystyle x_1 = 3\) and \(\displaystyle x_2 = -9\), and since we were able to find an interval for \(\displaystyle x\), \(\displaystyle -9\leqslant x \leqslant 7\), we see that both of these roots lie within the interval (and therefore both ought to be correct).

Here's the problem: \(\displaystyle x_1 = -9\) is an extraneous solution. Where did I go wrong? I have redone the algebra multiple times and even resorted to computational engines, and it still turns out that \(\displaystyle -9\) is a false root. Where is the flaw in my argument?