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- Thread starter Loki123
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I am not sure I understand why.sqrt(25) = 5 so nothing under the sqrt signs should be bigger than 25. OK?

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Hi Loki. We must always check answers obtained by squaring both sides of an equation because squaring can eliminate negative signs.Is there a way for me to discard 165 without going back to check the answers.

You'd correctly found that the domain of x is all numbers greater than or equal to 3. Therefore, both of your x-values 5 and 165 are valid candidates for solution.

The next step is to find corresponding y-values. Use the first given equation for that. You will get four (x,y) pairs. Two of those pairs may be eliminated because the first given equation tells us that y must be greater than or equal to zero.

Check the remaining two pairs using the following equation, and you'll find that only one of them works.

sqrt(2y^2 - 14) + y = 5

The square root of a Real number must be positive or zero.sqrt can only be positive

I am confused. Where did y come from???Hi Loki. We must always check answers obtained by squaring both sides of an equation because squaring can eliminate negative signs.

You'd correctly found that the domain of x is all numbers greater than or equal to 3. Therefore, both of your x-values 5 and 165 are valid candidates for solution.

The next step is to find corresponding y-values. Use the first given equation for that. You will get four (x,y) pairs. Two of those pairs may be eliminated because the first given equation tells us that y must be greater than or equal to zero.

Check the remaining two pairs using the following equation, and you'll find that only one of them works.

sqrt(2y^2 - 14) + y = 5

The square root of a Real number must be positive or zero.

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The right hand side is 5. The left hand side is the sum of square roots so each term is zero or more. If any of the square root term is more than 5, then the how left hand side will be more than 5.I am not sure I understand why.

So sqrt(2x-6)<= 5 and sqrt(x+4)<=5

That is 2x-6<=25 or

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Are you asking about potential y-values corresponding to x=5 and x=165? If so, you need to find them.I am confused. Where did y come from???

From the first given equation, you know that y is sqrt(x+4). Use substitution (x=5 and x=165) and evaluate.

Okay, I understand now how you got that, however, I am not sure why we have to set those conditions. I understand that they have to equal 5 combined so thats why, but why won't solving the equation normally take that into account on its own. For instance if I have (2x+2)+(3x+1)=13 I am not going to write 2x+2<=13 x<=11/2 or 3x+1<=13 x<=4...The right hand side is 5. The left hand side is the sum of square roots so each term is zero or more. If any of the square root term is more than 5, then the how left hand side will be more than 5.

So sqrt(2x-6)<= 5 and sqrt(x+4)<=5

That is 2x-6<=25 orx<=31/2andx+4<=25 orx<=21. So x<=31/2

Technically those conditions are true, but there is no reason for them... So why use them now? And don't say "it's because they are sqrts" because I still don't understand how that does anything...

I didn't use y. The second picture is the solution and the first one is my attempt. I didn't use y because frankly I don't see a reason to. If you are implying I need to insert a function y, please be more elaborate because so far I am lost.Are you asking about potential y-values corresponding to x=5 and x=165? If so, you need to find them.

From the first given equation, you know that y is sqrt(x+4). Use substitution (x=5 and x=165) and evaluate.

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Yes, that is obvious. Your work shows that you got to candidates x=5, x=165 and then stopped.The second picture is the solution and the first one is my attempt

What y values correspond to those potential x-values?

x = y^2 - 4

Substitute x=5 and solve for y. That yields candidates (5,3) and (5,-3).

Substitute x=165 and solve for y. That yields candidates (165,13) and (165,-13).

Check to see which candidates work in all given equations.

PS: You showed the solution, but you'd never posted the exercise. I had to guess which info in the solution was given in the exercise statement. Maybe I've guessed incorrectly. In future threads, please post exercise statements as given.

It's this one.Yes, that is obvious. Your work shows that you got to candidates x=5, x=165 and then stopped.

What y values correspond to those potential x-values?

x = y^2 - 4

Substitute x=5 and solve for y. That yields candidates (5,3) and (5,-3).

Substitute x=165 and solve for y. That yields candidates (165,13) and (165,-13).

Check to see which candidates work in all given equations.

PS: You showed the solution, but you'd never posted the exercise. I had to guess which info in the solution was given in the exercise statement. Maybe I've guessed incorrectly. In future threads, please post exercise statements as given.

I am sorry, but I don't understand why we need y values.

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I didn't use y. The second picture is the solution and the first one is my attempt. I didn't use y because frankly I don't see a reason to. If you are implying Ineedto insert a function y, please be more elaborate because so far I am lost.

It is notI am sorry, but I don't understand why weneedy values.

I would have done exactly what you did, because that is how textbooks I've taught from do it. Their method gave them a way to eliminate an extraneous solution; but I'm not sure it guarantees that

But even they haven't (in what you showed) promised that youIs there a way for me to discard 165 without going back to check the answers.

This is a good observation. What is happening in these problems is that each time you square, you are losing information about a sign; so in your work you are actually solving four equations simultaneously: [imath]\pm\sqrt{2x-6} \pm \sqrt{x+4}=5[/imath].For instance when I put 165 as x I get 18+13, it would work if it was -13 but sqrt can only be positive

It happens that 5 was the solution of [imath]+\sqrt{2x-6} + \sqrt{x+4}=5[/imath], and 165 was the solution of [imath]\sqrt{2x-6} - \sqrt{x+4}=5[/imath]. (The other two equations, with -+ and --, have no solutions.) I suspect that if they had let [imath]y=\sqrt{2x-6}[/imath], they might not have caught the extraneous solution, though I haven't bothered to try.

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We don't need to introduce any new variables. While guessing (from your image) what they'd given you, I'd thought some of the equations involving y had been provided in the exercise statement.I don't understand why we need y values.

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165 is not an answer. It does not satisfy the given equation.How is the answer 165 excluded?

[imath]\;[/imath]

I just tried it. You get 165 and 5 that way too.I suspect that if they had let y=2x−6y=\sqrt{2x-6}y=2x−6, they might not have caught the extraneous solution, though I haven't bothered to try.

I thought that I missed some condition and that that was the reason i got two answers.165 is not an answer. It does not satisfy the given equation.

[imath]\;[/imath]

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You did not get two answers. There is only one answer. (You'd gotten two results, which I call "candidates".)I thought that I missed some condition and that that was the reason i got two answers.

We often get extraneous solutions, after having squared an equation. That is why we need to check all results.

[imath]\;[/imath]

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You can add a very positive number to some number to sum to 5. You can add a very negative number to a number to sum to 5. In fact, the line x+y = 5 shows this.Okay, I understand now how you got that, however, I am not sure why we have to set those conditions. I understand that they have to equal 5 combined so thats why, but why won't solving the equation normally take that into account on its own. For instance if I have (2x+2)+(3x+1)=13 I am not going to write 2x+2<=13 x<=11/2 or 3x+1<=13 x<=4...

Technically those conditions are true, but there is no reason for them... So why use them now? And don't say "it's because they are sqrts" because I still don't understand how that does anything...

However, your problems does not allow this! The terms that you are adding are both non-negative. This restricts both terms to between 0 and 5 inclusive. Do you see this?

Basically you need to understand how to graph |x| + |y| = 5. What value can |x| be? How about |y|? Note that I did not ask what x and y can be!

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NO, there is no reason at all why 2x+2 <=13! 2x+2 can equal 113 as long as 3x+1 = -100.Okay, I understand now how you got that, however, I am not sure why we have to set those conditions. I understand that they have to equal 5 combined so thats why, but why won't solving the equation normally take that into account on its own. For instance if I have (2x+2)+(3x+1)=13 I am not going to write 2x+2<=13 x<=11/2 or 3x+1<=13 x<=4...

Technically those conditions are true, but there is no reason for them... So why use them now? And don't say "it's because they are sqrts" because I still don't understand how that does anything...

There is no reason at all to assume that both 2x+2 and 3x+1 are both positive or 0!

However, in your problem both terms MUST be positive or 0. You can't take the sqrt of a number and get less than 0. Don't you see the difference? You say that you understand but then give an example which shows you do not understand.

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