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Distributing into a square root

  1. Sep 25, 2012 #1
    Its been a while since I have taken any kind of math class, I am a bit rusty in general algebra. Can someone explain how I would multiply an equation like this

    (2x-1)sqrtof x-3x

    is it just like normal distribution? Would I just put the answer underneath the square root?
    sqrt2x^2-6x^2-x+3x?
     
  2. jcsd
  3. Sep 25, 2012 #2
    all of these things are equal:

    a*sqrt(b) = sqrt(a*a) * sqrt(b) = sqrt(a*a*b)
     
  4. Sep 25, 2012 #3
    To figure something like this out, try it with regular old numbers.

    For example [itex]5 \sqrt{4} = 5 * 2 = 10[/itex].

    But if you just put the 5 under the square root sign to make it sqrt(5*2) = sqrt(10), then that's not the same thing as 10 so you can't do that.

    Why not? Well, [itex]\sqrt{a^2b^2} = ab[/itex], right? That's because

    (ab)2 = a2b2.

    So, what's the fix? If we have 5 * sqrt(4) we can put the 5 under the radical by squaring it:

    [itex]5 \sqrt{4} = \sqrt{5^2*4} = \sqrt{100} = 10[/itex] as it should be.
     
    Last edited: Sep 25, 2012
  5. Sep 26, 2012 #4
    Ok so let me know if im the right track if I have (9y+1)sqrt 82
    i just square 9y+1 and put it under the square root with 82 and then times them both together?
     
  6. Sep 26, 2012 #5
    Yes, but now you have to be careful. If 9y+1 is negative, squaring it will lose information. So this depends on the context.

    In other words it is not always true that [itex]\sqrt{x^2} = x[/itex]. That's because the meaning of the square root symbol is the positive number that squares to what's under the radical. So if you start with x = -5, you'll end up introducing an error.

    Why do you want to put this expression under the radical? In general, doing so will change the meaning and introduce an error.
     
  7. Sep 26, 2012 #6
    Ahh I see well I am finding the area of a surface and I need to distribute this expression into the square root due to the formula I was given

    A= 2pi integral from a to b f(x)sqrt 1+ f(x) prime^2

    that is the forumula that I have to use
     
  8. Sep 27, 2012 #7

    HallsofIvy

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    I think you mean "f(x)sqrt(1+ f(x) prime^2)". Please use parentheses!

    [tex]\int_a^b f(x)\sqrt{1+ f'^2(x)}dx[/tex]

    Yes, you can write that as
    [tex]\int_a^b \sqrt{f^2(x)(1+ f'^2(x))}dx[/tex]

    Whether that is a good idea or not depends upon f.
     
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