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Equation: √(x+13) - √(7-x) = 2

sweatingbear

Member
May 3, 2013
91
We are solving the equation below for \(\displaystyle x \in \mathbb{R}\).

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

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?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
There is no flaw in your argument, and the "solution" $x=-9$ is indeed extraneous. You can see how it occurs if you put $x=-9$ in the original equation, which then becomes $\sqrt4 - \sqrt{16} = 2$. If you take the negative square root of $4$, and also the negative square root of $16$, then you get $(-2) - (-4) = 2$, which is correct. But it is conventional to use the notation $\sqrt{} $ to mean the positive square root, so this "solution" has to be rejected.

The only way to detect such extraneous solutions in problems like this is to check them in the original problem, which is just what you did.
 

sweatingbear

Member
May 3, 2013
91
Yes but an idiosyncrasy of mine is that I do not think testing solutions is as elegant as being able to state an interval for the variable in question and thereby rule out incorrect solutions.

I do not understand why we are not able to restrict \(\displaystyle x\) into an interval with which we can discard extraneous solutions? It has worked for me just fine in other problems.
 

kaliprasad

Well-known member
Mar 31, 2013
1,309
Yes but an idiosyncrasy of mine is that I do not think testing solutions is as elegant as being able to state an interval for the variable in question and thereby rule out incorrect solutions.

I do not understand why we are not able to restrict \(\displaystyle x\) into an interval with which we can discard extraneous solutions? It has worked for me just fine in other problems.
The problem comes because of squaring

x- y = z when squared we get (x-y)^2 = z^2 and this includes solution of x-y = - z. so extraneous solutions occur for these cases
 

sweatingbear

Member
May 3, 2013
91
The problem comes because of squaring

x- y = z when squared we get (x-y)^2 = z^2 and this includes solution of x-y = - z. so extraneous solutions occur for these cases
Yes I already understand the why of the origin of extraneous solutions, I just do not understand why I – in this case – was unable to state an interval for \(\displaystyle x\) from which I could have ruled out the extraneous root \(\displaystyle x = -9\). It has worked for me before, why would it not now? Certainly, this is equation is not some kind of a special case where this method fails?
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
I just do not understand why I – in this case – was unable to state an interval for \(\displaystyle x\) from which I could have ruled out the extraneous root \(\displaystyle x = -9\).
As kaliprasad said, the additional interval restriction on $x$ occurs during the transition from \[ \sqrt{x+13} - \sqrt{7-x} = 2\] to \[\left(\sqrt{x+13}-\sqrt{7-x}\right)^2=4\] This restriction is \[\sqrt{x+13} - \sqrt{7-x}\ge0 \iff \sqrt{x+13}\ge \sqrt{7-x} \iff -3\le x\le 7\]

It is important to understand that transformation we apply to equalities to solve them are not some magic operations given to us from above. If that were so, then our mathematics would resemble that of ancient Babylonians or Egyptians, which, I heard, was based on algorithms and not on proofs. As it is, the complete solution of an equation $E(x)=0$ must represent (or give rise to) a proof of the equivalence \[E(x)=0\iff x=x_1\text{ or }x=x_2\text{ or }\dots\text{ for all }x\]where $x_1,x_2,\dots$ are alleged roots. The easiest way to achieve this is to transform the equation using equivalences. Theorems needed to solve the original equation include
\begin{gather*}x=y\iff x^2=y^2\text{ and }x\text{ and }y\text{ have the same sign}\\ \left(\sqrt{x}\right)^2=x\text{ for }x\ge0\\ \sqrt{\strut x}\sqrt{\strut y}=\sqrt{\strut xy}\text{ for }x\ge0,y\ge0\\ \sqrt{x}\ge\sqrt{y}\iff x\ge y\ge0\end{gather*} The first of these theorems generates the additional requirement $x\ge-3$.
 

sweatingbear

Member
May 3, 2013
91
@Evgeny.Makarov: Excellent answer, just the type I was looking for. Thanks a heap!