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Solving inequalities with radicals

I have a comment and question to make about some recent posts solving inequalities with a radical in them.

Consider the inequality:

$\displaystyle x - 2 \geq \sqrt{13 - 4x}$

I think we would all agree that

$\displaystyle 13 - 4x \geq 0 \implies x \leq \frac{13}{4}$

as a reasonable condition on the solutions to this inequality.

But from there I have seen several instances of the following procedure:

$\displaystyle (x - 2)^2 \geq 13 - 4x$

$\displaystyle x^2 - 4x + 4 \geq 13 - 4x$

$\displaystyle x^2 - 9 \geq 0$

$\displaystyle (x + 3)(x - 3) \geq 0$

Thus the solution set for x would be $\displaystyle (-\infty, -3] \cup \left [ 3, \frac{14}{3} \right ]$.

But if you look at the graph of $\displaystyle y = (x - 2) - \sqrt{13 - 4x}$ you will see that the only solutions are for $\displaystyle \left [ 3, \frac{14}{3} \right ]$.

So why does this happen? The only thing I can think of is that when $\displaystyle x - 2 \leq 0$ then by squaring both sides of the inequality we are multiplying both sides by a negative number, switching the $\displaystyle \geq$ to $\displaystyle \leq$. So the $\displaystyle (-\infty, -3]$ interval is discarded from our previous solution since this interval makes $\displaystyle x - 2 < 0$.

The new inequality gives

$\displaystyle (x - 2)^2 \leq 13 - 4x$

$\displaystyle x^2 - 4x + 4 \leq 13 - 4x$

$\displaystyle x^2 - 9 \leq 0$

$\displaystyle (x + 3)(x - 3) \leq 0$

The solution set to this is $\displaystyle [-3, 3] $. But this solution does not contribute at all to the solution set. What has gone wrong?

-Dan