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Question about squares/square roots

  1. Aug 27, 2006 #1
    I have a feeling the answer to my question is pretty simple, but I couldn't come up with an answer, so I'm going to ask it here.

    Say you have an equation which looks like this:

    y^2 = x + 4

    Assuming you want to rearrange the equation to solve for y, you would do this:

    y = square root(x + 4)

    I understand what is being done here, however I don't fully understand the reasoning behind it. I know that you are taking the square root of both sides, but I wondered why you couldn't do this another way, the way I thought of, is as follows:

    y = square root(x) + square root(4)

    Now, I know that these two ways are not the same, but I wondered why they aren't the same, and what I wondered about even more, is how you know to choose one way over the other. If the equation was set up like this:

    2y = x + 4

    Then you could do this:

    y = (x + 4)/2

    Or you could do this, and still get the same result:

    y = x/2 + 4/2

    Dividing each term individually and then adding the results has the same effect as adding the terms, and then dividing the result. If you can use two different methods in this case, and get the same result, then how do you know which way to use in the other case(with squares/square roots)? This is my main question, but I'm also wondering why the answers end up different in the first place.
     
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  3. Aug 27, 2006 #2

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    What you always do when you solve these kinds of algebraic equations is to do one thing to one side of the equal sign and the same thing to the other. This way you end up with another equality. So, for example, you might start with:

    [tex]y^2-x-4=0[/tex]

    now you add x+4 to both sides:

    [tex]y^2=x+4[/tex]

    now you take the square root of both sides:

    [tex]y=\pm \sqrt{x+4}[/tex]

    (by the way, the plus/minus is an inherent ambiguity of the square root operation. it doesn't affect the argument).

    Now, if you're asking why [itex] \sqrt{a+b} \neq \sqrt{a}+\sqrt{b}[/itex], just square both sides to see. sqrt is just a function, and not all functions are linear, in the sense that f(a+b)=f(a)+f(b) (in fact, the overwhelming majority are not). Other examples of nonlinear functions are [itex] (a+b)^2 \neq a^2+b^2[/itex] and [itex] \sin(a+b) \neq \sin(a)+\sin(b)[/itex]. f(x)=x/2 is linear, however, and so [itex](a+b)/2=a/2+b/2[/itex], and these can be interchanged in any equation.
     
    Last edited: Aug 27, 2006
  4. Aug 27, 2006 #3
    Ok, I think I understand a bit better now, because the square root/square function is not a linear function, the two methods won't cause the function to end up the same, right?

    But why do you know to use one method over the other? I know that the methods don't give the same answer, but I'm wondering exactly why you choose one method over the other. Why do we know, that in the case of y^2 = x + 4, that, to simplify for y, we use the square root of x + 4, instead of the square root of x plus the square root of 4? Why do we know that this is the correct answer?
     
  5. Aug 27, 2006 #4

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    Like I said, you apply these operations to each side as a whole. Since each side is equal by assumption, the square root of one side equals the square root of the other, just like the nth power of one equals that of the other, or one side divided by 34 equals the other divided by 34. This is why algebra works. Your confusion is coming because you usually skip the middle step in this sequence:

    [tex]2y=x+4[/tex]

    [tex]y=(x+4)/2[/tex]

    [tex]y=x/2+4/2[/tex]

    It's fine that you skip this step, and you'd be crazy to write it out every time. But you have to keep in mind that the only reason you can go from the 2nd to the 3rd line is because the function f(x)=x/2 is linear.
     
  6. Aug 27, 2006 #5
    Oh, I think I get it now. In my example:

    y^2 = x + 4

    If you took the square root of x and the square root of 4 individually, and then added them, the sides wouldn't be equal, right? Is there a way to test this with an equation? I can't think of a way to do so right now.
     
  7. Aug 27, 2006 #6

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    You can show [itex]\sqrt{a+b} \neq \sqrt{a}+\sqrt{b}[/itex] by squaring both sides. Of course, there are some cases when this isn't true, like if a=5, b=0. But the point is that in general there isn't an equality.
     
  8. Aug 27, 2006 #7
    Ok, so if I squared [itex]\sqrt{a+b}[/itex], I'd end up with a + b, right? But if I squared [itex]\sqrt{a}+\sqrt{b}[/itex] I'd end up with [itex]\sqrt{a}+\sqrt{b} * \sqrt{a}+\sqrt{b}[/itex], right? In that case, you could simplify to [itex]a + \sqrt{a} * \sqrt{b} + b[/itex], right?
     
    Last edited: Aug 27, 2006
  9. Aug 27, 2006 #8

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    Come on, you should be able to square something on your own. What you have there is wrong.
     
  10. Aug 27, 2006 #9
    Was the first one done right, at least?

    I'll go over the second one again.

    [itex]\sqrt{a}+\sqrt{b}[/itex]
    [itex]=(\sqrt{a}+\sqrt{b})^2[/itex]
    [itex]=a + 2(\sqrt{a})(\sqrt{b}) + b[/itex]

    Did I do it right this time?
     
    Last edited: Aug 27, 2006
  11. Aug 27, 2006 #10

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    Yea, they're both right now.
     
  12. Aug 27, 2006 #11
    Ok, I think I understand this much pretty well now, I was also wondering... what about addition/subtraction? In a case like this:

    y + 4 = x + 5

    You wouldn't do this:

    y = x - 4 + 5 - 4

    You would do this:

    y = x + 5 - 4

    Why do you do the first method instead the second? Is this related to the fact that terms themselves are separated by plus/minus signs, or is this related to your earlier advice? Or is it a combination of both?
     
    Last edited: Aug 27, 2006
  13. Aug 27, 2006 #12

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    I don't understand what you're asking. Could you use an example?
     
  14. Aug 27, 2006 #13
    I editted my post to include an example, if it helps. Or did you already see how I editted it before making your post? In that case, I don't know how I can clarify my post any more.
     
  15. Aug 27, 2006 #14

    matt grime

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    There is at least one mistake in post 11. The two things you claim are different are the same.

    if y+5=x+4 then y=x-1, just take 5 off from both sides.
     
  16. Aug 27, 2006 #15
    Woops, I just editted that post, thanks for pointing that out, maybe now it will be a little more clear as to what I was asking.
     
  17. Aug 27, 2006 #16

    matt grime

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    Well, you have a slight contradiction, still.

    You would do it the 'proper' way because that is 'how numbers work', and it is no great mystery.

    if y+4=x+5, then y=x+1, why would you subtract 4 twice from the right hand side when you only subtracted it once from the left hand side?
     
  18. Aug 27, 2006 #17
    A slight contradiction? What do you mean, exactly? Where is the contradiction in my post? I can clarify what I posted, if you want.

    Anyway, I think i see what you're doing now, 4 is only subtracted once on the left side, so why would you do it twice on the right side? My thining was that, with multipltication/division, you can multiply/divide by each term, and still achieve the same answer you would if you multiplied/divided a whole side first, and then added the resulting terms, so why can't you apply this to addition/subtraction? That's what I'm wondering, but I think I'm missing something.
     
  19. Aug 27, 2006 #18

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    That's because multiplication distributes over addition, that is, a*(b+c)=(a*b)+(a*c). It is not true that a+(b+c)=(a+b)+(a+c). Remember, ultimately all these symbols are supposed to stand in for real numbers. Plug in a few numbers for a,b,c in the above two expressions to see why the first is true and the second isn't.
     
  20. Aug 27, 2006 #19

    matt grime

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    So you want, in some sense, (a+b)+c to be the same as (a+b)+(b+c). Well, if we did that then there would be no way to say what a number is.

    Since 1=1+0= 1+(0+0), then in this system, 1+(0+0)=(0+1)+(0+1)=1+1=2. So doing it this way 1+1=2 (and hence 3, 4, and indeed any other integer).


    (The contradiction referred to the fact that you said the wrong one was the correct one, at one point, then swapped it over again later).
     
  21. Aug 27, 2006 #20
    I know that a*(b+c)=(a*b)+(a*c), and that it is not true that a+(b+c)=(a+b)+(a+c), but I was wondering why we don't add/subtract to each term in the first place, instead of just adding it to the side of the equation. But I guess we use this method, because of what you said earlier, something similar to:

    a*b = c+d
    a = (c+d)/b
    a = c/b+d/b

    The second step being one which is frequently skipped over, which lead to my confusion. What I was wondering, though, was why the second step and third step can't be reversed when doing non-muliplication/division operations(such as squaring/taking the square roots), and yet it can be done when doing multiplication/division.

    Then I came up with an example to help me understand this:

    [itex]\sqrt{x}=5+5[/itex]
    [itex]x=(5+5)^2[/itex]
    [itex]x=10^2[/itex]
    [itex]x=100[/itex]

    If you substitute 100 back in, you get the following:

    [itex]\sqrt{100}=5+5[/itex]
    [itex]10=10[/itex]

    Everything works out this way, however, if you do it the other way, you don't get the right answer:

    [itex]\sqrt{x}=5+5[/itex]
    [itex]x=5^2+5^2[/itex]
    [itex]x=25+25[/itex]
    [itex]x=50[/itex]

    When you substitute 50 for x, this is what happens:

    [itex]\sqrt{50}=5+5[/itex]
    [itex]7.1=10[/itex]

    You don't get the right answer this way. Though, I don't really like my example, because 5 and 5 are like terms, and could easily have been combined to get 10 in the first step, before showing how to properly take the square root of that side. Could someone else show, or at least sugggest another example, similar to this one?
     
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