SweatingBear
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We are solving the equation below for $$x \in \mathbb{R}$$.
$$\sqrt{x+13} - \sqrt{7-x} = 2 \, .$$
If $$x \in \mathbb{R}$$ then we must require that
$$\begin{cases}
x+13 \geqslant 0 & \iff & x \geqslant -13 \\
7-x \geqslant 0 & \iff & x \leqslant 7
\end{cases}$$
Moreover, if we rewrite equation as $$\sqrt{x+13} = 2 + \sqrt{7-x}$$, the right-hand side must in fact be greater than or equal to zero (since the output of $$\sqrt{x+13}$$ is greater than or equal to zero). Thus
$$2 + \sqrt{7-x} \geqslant 0 \iff \sqrt{7-x} \geqslant -2 \, .$$
but $$\sqrt{7-x} \geqslant -2 \ , \ \forall x \in \mathbb{R}$$ so no further restriction on $$x$$ could be derived.
We could rewrite the equation as $$\sqrt{7-x} = \sqrt{x+13} - 2$$ and apply a similar reasoning about the output of the square root function. We must require $$\sqrt{x+13} - 2 \geqslant 0$$.
$$\sqrt{x+13} -2 \geqslant 0 \iff \sqrt{x+13} \geqslant 2 \iff x+13 \geqslant 4 \iff x \geqslant -9 \, .$$
So, the restrictions we have on $$x$$ thus far are
$$(x \geqslant -13) \wedge (x \leqslant 7) \wedge (x\geqslant -9) \implies -9 \leqslant x \leqslant 7 \, .$$
Solving the aforementioned equation by squaring a few times and using the quadratic formula yields the roots $$x_1 = 3$$ and $$x_2 = -9$$, and since we were able to find an interval for $$x$$, $$-9\leqslant x \leqslant 7$$, we see that both of these roots lie within the interval (and therefore both ought to be correct).
Here's the problem: $$x_1 = -9$$ 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 $$-9$$ is a false root. Where is the flaw in my argument?
$$\sqrt{x+13} - \sqrt{7-x} = 2 \, .$$
If $$x \in \mathbb{R}$$ then we must require that
$$\begin{cases}
x+13 \geqslant 0 & \iff & x \geqslant -13 \\
7-x \geqslant 0 & \iff & x \leqslant 7
\end{cases}$$
Moreover, if we rewrite equation as $$\sqrt{x+13} = 2 + \sqrt{7-x}$$, the right-hand side must in fact be greater than or equal to zero (since the output of $$\sqrt{x+13}$$ is greater than or equal to zero). Thus
$$2 + \sqrt{7-x} \geqslant 0 \iff \sqrt{7-x} \geqslant -2 \, .$$
but $$\sqrt{7-x} \geqslant -2 \ , \ \forall x \in \mathbb{R}$$ so no further restriction on $$x$$ could be derived.
We could rewrite the equation as $$\sqrt{7-x} = \sqrt{x+13} - 2$$ and apply a similar reasoning about the output of the square root function. We must require $$\sqrt{x+13} - 2 \geqslant 0$$.
$$\sqrt{x+13} -2 \geqslant 0 \iff \sqrt{x+13} \geqslant 2 \iff x+13 \geqslant 4 \iff x \geqslant -9 \, .$$
So, the restrictions we have on $$x$$ thus far are
$$(x \geqslant -13) \wedge (x \leqslant 7) \wedge (x\geqslant -9) \implies -9 \leqslant x \leqslant 7 \, .$$
Solving the aforementioned equation by squaring a few times and using the quadratic formula yields the roots $$x_1 = 3$$ and $$x_2 = -9$$, and since we were able to find an interval for $$x$$, $$-9\leqslant x \leqslant 7$$, we see that both of these roots lie within the interval (and therefore both ought to be correct).
Here's the problem: $$x_1 = -9$$ 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 $$-9$$ is a false root. Where is the flaw in my argument?