Restrictions in Radical Equations

[idtoms]

Hi there,

I have a question about restrictions in radical equations. If I have the equation sqrt(x^2+3)=2x , should I say there are no restrictions since there are no values that cause either side to be undefined? Or should I say that x should be greater than or equal to zero since the principal square root is going to be positive?

Help please!!

Thanks,

Ian

 

[tkhunny]

Hi there,

I have a question about restrictions in radical equations. If I have the equation sqrt(x^2+3)=2x , should I say there are no restrictions since there are no values that cause either side to be undefined? Or should I say that x should be greater than or equal to zero since the principal square root is going to be positive?

Help please!!

Thanks,

Ian

That's very good. I'm glad you are concerned about this issue.
You should say what it is. There are multiple types of restrictions. Consider them all.

Here is a restriction type based on the failure of the square root if it is given a negative number. More of a Domain restriction.
Inside the radical. \(\displaystyle x^{2} + 3 \ge 0\). There is no value of x that violates this.

Here is a restriction type that emerges because you simply can't produce every Real Number. More of a Range restriction.
The radical, itself. \(\displaystyle \sqrt{x^{2}+3} \ge \sqrt{3}\), so \(\displaystyle 2x \ge \sqrt{3} \rightarrow x \ge \dfrac{\sqrt{3}}{2}\).


Don't say \(\displaystyle 0\) when you should say \(\displaystyle \sqrt{3}\). Not all restrictions are obvious. That's why you still need to check your results. In this case, you may have gotten x = 1 as a solution. That wouldn't work, even though you thought it was allowed with \(\displaystyle x \ge 0\).

 

[JeffM]

That's very good. I'm glad you are concerned about this issue.
You should say what it is. There are multiple types of restrictions. Consider them all.

Here is a restriction type based on the failure of the square root if it is given a negative number. More of a Domain restriction.
Inside the radical. \(\displaystyle x^{2} + 3 \ge 0\). There is no value of x that violates this.

Here is a restriction type that emerges because you simply can't produce every Real Number. More of a Range restriction.
The radical, itself. \(\displaystyle \sqrt{x^{2}+3} \ge \sqrt{3}\), so \(\displaystyle 2x \ge \sqrt{3} \rightarrow x \ge \dfrac{\sqrt{3}}{2}\).
Don't say \(\displaystyle 0\) when you should say \(\displaystyle \sqrt{3}\). Not all restrictions are obvious. That's why you still need to check your results.

EDITED POST

I wholeheartedly agree that considering implied restrictions on the original problem is excellent technique.
I further agree that saying \(\displaystyle \sqrt{x^2 + 3} > 0\) does not adequately express \(\displaystyle \sqrt{x^2 + 3} \ge \sqrt{3}.\)

So this was an excellent answer up to here.

And then, as a former bridge partner used to say to me, a cow flew by: x = 1 is a valid answer because

\(\displaystyle 3 < 4 \implies \sqrt{3} < \sqrt{4} \implies \sqrt{3} < 2 \implies \dfrac{\sqrt{3}}{2} < 1.\)

I have seen my share of flying cows so I sympathize.

 

[tkhunny]

I am confused by this answer.


Thus, x = 1 seems to me to be a perfectly valid answer because

\(\displaystyle 3 < 4 \implies \sqrt{3} < \sqrt{4} \implies \sqrt{3} < 2 \implies \dfrac{\sqrt{3}}{2} < 1.\)

So I am lost here:

Simple. That's called forgetting the "/2". My bad. Thanks for pointing that out.

OP: Whoops, make that x = 1/2 as a possibly misleading result. (or x = 3/4 or x = 1/5 or etc.) x = 1 is fine.

 

[JeffM]

Simple. That's called forgetting the "/2". My bad. Thanks for pointing that out.

OP: Whoops, make that x = 1/2 as a possibly misleading result. (or x = 3/4 or x = 1/5 or etc.) x = 1 is fine.

Oh. I was afraid there was some subtle thing that I had not seen but that you had not explained. I shall review my previous post and edit accordingly.

 

[idtoms]

Thanks so much for the clear reply! So basically, I look for "domain-type" restrictions within the radical. Next I should look at any "output" features of the radical and build/solve an inequality. The full restriction will come from all of the considerations combined?

 

[tkhunny]

Thanks so much for the clear reply! So basically, I look for "domain-type" restrictions within the radical. Next I should look at any "output" features of the radical and build/solve an inequality. The full restriction will come from all of the considerations combined?

Notice the idea of "look for". You may miss things. Some are more subtle than others. You may forget to divide by 2, somewhere. Always check you results.

 

[Dr.Peterson]

I have a question about restrictions in radical equations. If I have the equation sqrt(x^2+3)=2x , should I say there are no restrictions since there are no values that cause either side to be undefined? Or should I say that x should be greater than or equal to zero since the principal square root is going to be positive?

I've been wanting to ask you about the context of your question. Are you asking because your teacher wants you to state restrictions, or for some other reason? In the former case, you would want to go by the teacher's intention: if they mean just restrictions on the domain (which I think may be common), then that is all you need to do. On the other hand, if you just want to avoid giving invalid solutions, it can be easier just to check each solution you find to see if it is valid.

In this case, as you presumably have found, when you solve by squaring both sides, you get 3 = 3x^2, so x can be either 1 or -1. It can be useful to have already noticed that x can't be negative, so that -1 will not be a valid solution; but you will find that out when you check in the original equation, which you should do anyway.

So finding precise restrictions on domain and range can be overkill, if you are just trying to solve the equation. There are other situations (such as solving a formula, where the result is not just a number you can check, or finding an inverse function) where the details can be important.

Actually, the ultimate "restriction" is that x can only be 1, because no other number satisfies the equation!

 

[idtoms]

I've been wanting to ask you about the context of your question. Are you asking because your teacher wants you to state restrictions, or for some other reason? In the former case, you would want to go by the teacher's intention: if they mean just restrictions on the domain (which I think may be common), then that is all you need to do. On the other hand, if you just want to avoid giving invalid solutions, it can be easier just to check each solution you find to see if it is valid.

In this case, as you presumably have found, when you solve by squaring both sides, you get 3 = 3x^2, so x can be either 1 or -1. It can be useful to have already noticed that x can't be negative, so that -1 will not be a valid solution; but you will find that out when you check in the original equation, which you should do anyway.

So finding precise restrictions on domain and range can be overkill, if you are just trying to solve the equation. There are other situations (such as solving a formula, where the result is not just a number you can check, or finding an inverse function) where the details can be important.

Actually, the ultimate "restriction" is that x can only be 1, because no other number satisfies the equation!

The reason I asked is because something similar came up in an exam I just wrote and I won't get the exam back so I wasn't sure what the right answer is. The actual question in the exam was to determine the restrictions on this equation:

9-sqrt(9+x^2)=x

I feel like I could have solved it fine (I think the answer is 4), but I wasn't sure about the restrictions. I like the idea of the solution as the ultimate restriction though. Shoulda gone with that.

 

[idtoms]

I've been wanting to ask you about the context of your question. Are you asking because your teacher wants you to state restrictions, or for some other reason? In the former case, you would want to go by the teacher's intention: if they mean just restrictions on the domain (which I think may be common), then that is all you need to do. On the other hand, if you just want to avoid giving invalid solutions, it can be easier just to check each solution you find to see if it is valid.

I tried to reply to this on Friday, but I think I may have a lag in my post since I'm a newbie.

I'm asking because a similar question came up on an exam that I'm not going to get back and it got me thinking about these restrictions. The question was to state the restrictions on the equation 9-sqrt(9+x^2)=x . The options were:

(a) no restrictions
(b) -9<=x<=9
(c) -6<=x<=6
(d) x not equal to 9 or -9

In class we had done "domain-style" restrictions, so I was tempted by "no restrictions". But then I thought that since x= "9 minus something", that it couldn't be "no restrictions". So I chose (b), because it had a <=9 in it. After reading the explanation in this thread, I can see that I probably should have chosen x<=6 (though I don't know where that -6 came from). Or maybe I should have chosen "no restrictions"? That's what most of my friends put.

I can figure out that the answer is 4, and I now think the restriction should be x<=6. But it's all kind of got my head spinning about the whole idea of restrictions in an equation like this. I guess my big questions are:

(1) If I do a perfect job of finding all restrictions, will I be able to figure out which roots are extraneous and which actually work without checking in the original? (I know I SHOULD check no matter what) Or are there some I'll miss even with a good set of restrictions?

(2) If I can solve the equation and check in the original, why is it important to find restrictions in these equations at all?

(3) What is actually meant by a restriction in a one variable equation? Is it just all the stuff I can tell about the solution without actually solving? Or is it just the stuff that breaks domain rules?