# Math Help - Solving inequalities with radicals

1. ## 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:
$x - 2 \geq \sqrt{13 - 4x}$

I think we would all agree that
$13 - 4x \geq 0 \implies x \leq \frac{13}{4}$
as a reasonable condition on the solutions to this inequality.

But from there I have seen several instances of the following procedure:
$(x - 2)^2 \geq 13 - 4x$

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

$x^2 - 9 \geq 0$

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

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

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

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

The new inequality gives
$(x - 2)^2 \leq 13 - 4x$

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

$x^2 - 9 \leq 0$

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

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

-Dan

2. Originally Posted by janvdl
It seems that as soon as we square both sides, we lose some of the solutions. Strange.
Really? What if the poster had posted $2 - x \geq \sqrt{13 - 4x}$??
After squaring.....Voila! the same equation!!

So what is the reason?

You have introduced an additional solution by squaring!

Now $a^2 \geq b^2 \Rightarrow (a-b)(a+b) \geq 0$,
So generally we choose an x such that x > y and claim x^2 > y^2(implicitly assuming x and y are positive)

In topsquarks problem, we implicitly assumed (x-2) is positive while doing manipulations. So the final solution is $[2, \infty) \cap (-\infty, -3] \cup \left [ 3, \frac{14}{3} \right ] = \left [ 3, \frac{14}{3} \right ]$.

3. Originally Posted by janvdl
In this specific abovementioned equation we lose solutions.
Which solutions did we lose?
I think you mean $[-\infty, -3]$. In that case your math is telling you, it is the solution to $2 - x \geq \sqrt{13 - 4x}$.
The point is we never lost any solutions, but only added more while squaring. So if we carefully remember all the places where we did this, we can finally write the right answer

4. Originally Posted by janvdl
So -4; -5; -6; -7; etc are not solutions as indicated in the graph i posted?

I don't understand. Why aren't they?
No.... janvdl, I am saying they are not the solutions to the original problem, that is $x - 2 \geq \sqrt{13 - 4x}$. But the moment you square it, {-3,-4,-5...} is a solution. Sub and check!

What I am trying to explain is that the moment you square it, the problems $x - 2 \geq \sqrt{13 - 4x}$ and $2 - x \geq \sqrt{13 - 4x}$ are virtually indistinguishable. And your subsequent math yields you $(-\infty, -3] \cup \left [ 3, \frac{14}{3} \right ]$. But then at this step you should ask yourself, which problem are you solving. So if it is the former then $\left [ 3, \frac{14}{3} \right ]$ is the solution. If it is the other one, $(-\infty, -3]$ is the solution

5. Originally Posted by Isomorphism
Really? What if the poster had posted $2 - x \geq \sqrt{13 - 4x}$??
After squaring.....Voila! the same equation!!

So what is the reason?

You have introduced an additional solution by squaring!

Now $a^2 \geq b^2 \Rightarrow (a-b)(a+b) \geq 0$,
So generally we choose an x such that x > y and claim x^2 > y^2(implicitly assuming x and y are positive)

In topsquarks problem, we implicitly assumed (x-2) is positive while doing manipulations. So the final solution is $[2, \infty) \cap (-\infty, -3] \cup \left [ 3, \frac{14}{3} \right ] = \left [ 3, \frac{14}{3} \right ]$.
You are correct. I figured this out over lunch in a slightly different way:

$\sqrt{13 - 4x}$ is defined to be positive for all allowed real values of x. So $x - 2 \geq = 0 \implies x \geq 2$. So the solution is, as you say, the intersection of the two sets.

-Dan