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Math Help - Help with Surds

  1. #1
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    Help with Surds

    Hi, I basically need some help with Surds and understand them and also general algebra. Basically the first question of mine is what is about simplifying:


    (Ill replace Square Root icon with a '*' as I cant find the icon anywhere to copy and paste)

    1. *1 + *2 + *4 + *8

    I originally thought that it would be just adding the numbers together to get *15 but then realised that isnt possible. Could someone please explain what exactly I have to do, not just give me the answer so I know how to do it in the future.

    Also what would x(squared)+x-2 be? I know it would start as:

    (x+ )(x- ) but dont know what goes in the gaps?

    Thanks a lot.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Yppolitia View Post
    (Ill replace Square Root icon with a '*' as I cant find the icon anywhere to copy and paste)

    *1 + *2 + *4 + *8
    This is really two separate questions, so I'll answer them that way. I'm also going to rewrite your problem as:
    sqrt(1) + sqrt(2) + sqrt(4) + sqrt(8)

    Now, we know sqrt(1) and sqrt(4) are 1 and 2 respectively. And sqrt(2) is, well, sqrt(2). We can't really do anything about this one. So we've got:
    sqrt(1) + sqrt(2) + sqrt(4) + sqrt(8) = 1 + sqrt(2) + 2 + sqrt(8) = 3 + sqrt(2) + sqrt(8)

    What about sqrt(8)? Well, 8 = 2*4 = 2*(2^2). So
    sqrt(8) = sqrt[2*(2^2)] = sqrt(2)*sqrt[(2^2)]

    But sqrt[(2^2)] = 2 since the square root and the ^2 undo each other. (Or you could simply say sqrt[(2^2)] = sqrt(4) = 2.) So
    sqrt(8) = sqrt[2*(2^2)] = sqrt(2)*sqrt[(2^2)] = sqrt(2)*2 = 2*sqrt(2)

    So
    sqrt(1) + sqrt(2) + sqrt(4) + sqrt(8) = 3 + sqrt(2) + sqrt(8) = 3 + sqrt(2) + 2*sqrt(2)
    = 3 + 3*sqrt(2)

    -Dan
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  3. #3
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    Hello, Yppolitia!

    I basically need some help with Surds and understand them.
    . . . . . . . . . _ . . ._ . . ._ . . _
    1) Simplify: √1 + √2 + √4 + √8

    . . . . . . . . . . . . _ . . . . . _ . . . . . _ . . . ._
    We know that: .√1 = 1, . √4 = 2, . √8 = 2√2
    . . . . . . . . . . . . . _ . . . . . . _ . . . . . . . . _
    So we have: .1 + √2 + 2 + 2√2 . = . 3 + 3√2



    Also: . x² + x - 2 .= .(x + .)(x - .)

    We know that the product of the two numbers is -2.
    There are only two choices: 1 and -2 . . . 2 and -1

    Try both and see which one give us the original trinomial.

    . . (x + 1)(x - 2) .= .x² - 2x + x - 2 .= .- x - 2 . . . no

    . . (x + 2)(x - 1) .= .x² - x + 2x - 2 .= .x² + x - 2 . . . yes!

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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Yppolitia View Post
    Also what would x(squared)+x-2 be? I know it would start as:

    (x+ )(x- ) but dont know what goes in the gaps?
    I'm going to do a different problem, then get to yours. The reason is that the method I'm going to show you is better demonstrated on a more complicated problem. This method is called the "ac method."

    Factor: x^2 - x - 6

    What we wish to do is first list all the factors of -6:
    -6, 1
    -3, 2
    -2, 3
    -1, 6
    1, -6
    2, -3
    3, -2
    6, -1

    Now, which of these combinations add to form the coefficient of the linear term? (Which is -1 in this case.) The answer is that there are two pairs listed that sums to -1: -3, 2 and 2, -3. It doesn't matter which one we pick, so I'll arbitrarily pick the first one.

    So what I'm going to do is rewrite the linear term in the original problem as -x = -3x + 2x:
    x^2 - x - 6 = x^2 - 3x + 2x - 6

    Now I'm going to use the method "factoring by grouping."
    x^2 - 3x + 2x - 6
    (x^2 - 3x) + (2x - 6)
    x(x - 3) + 2(x - 3)
    (x + 2)(x - 3).

    I'll leave it to you to check that if we had picked the 2, -3 pair that we get the same answer.

    The only thing to add if we have the more general problem of factoring
    ax^2 + bx + c
    is that instead of factoring "c" like we did above, we are going to factor a*c. (Hence the name "ac method.") The rest is identical with what I've shown you above.

    So, on to your problem:
    x^2+x-2

    List all the factors of 1*-2 = -2 (a = 1, c = -2 here):
    -2, 1
    -1, 2
    1, -2
    2, -1
    Which of these pairs add up to the coefficient of the linear term in the problem, which is 1 in this case? The answer is both -1, 2 and 2, -1 do. I'll again arbitrarily pick the first pair. Again it doesn't matter which one we pick.

    So I'm going to write x = -x + 2x in the original.
    x^2 + x - 2
    x^2 - x + 2x - 2
    (x^2 - x) + (2x - 2)
    x(x - 1) + 2(x - 1)
    (x + 2)(x - 1).

    Now, I'll admit that in this case the answer would have been easily gotten to by guessing:
    x^2 + x - 2 = (x + _)(x + _)
    The only possible numbers that can go in the slots are the factors of -2 as we listed earlier, so there are only 4 possibilities to try out.

    -Dan
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  5. #5
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    Great, thanks a lot. Sorry to be greedy but if you could help me answer these two other questions that would be very helpful

    What would x(squared)+x-2 be? I know it would start as:

    (x+ )(x- ) but dont know what goes in the gaps?

    And I need to reduce the following root to the form a(sqrt)b where 'a' is an integer and 'b' in as small as possible:

    (sqrt)50 = 2(sqrt)25 = 10(sqrt)5 (Is that right?)

    If that is right how would I do (sqrt)99 and (sqrt)1000.


    Really sorry for all these questions but my teachers is pretty arrogant and doesnt seem to like helping me.
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Yppolitia View Post
    What would x(squared)+x-2 be? I know it would start as:

    (x+ )(x- ) but dont know what goes in the gaps?
    Ummm...both Soroban and I went over this. Are you saying you are still having problems with it?

    Quote Originally Posted by Yppolitia View Post
    And I need to reduce the following root to the form a(sqrt)b where 'a' is an integer and 'b' in as small as possible:

    (sqrt)50 = 2(sqrt)25 = 10(sqrt)5 (Is that right?)
    Yes, that is perfect!

    Quote Originally Posted by Yppolitia View Post
    If that is right how would I do (sqrt)99 and (sqrt)1000.
    sqrt(99) = sqrt(9*11) = sqrt(9)*sqrt(11) = 3*sqrt(11)

    sqrt(1000) = sqrt(100*10) = sqrt(100)*sqrt(10) = 10*sqrt(10)

    -Dan
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  7. #7
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    Quote Originally Posted by Yppolitia View Post
    ...
    And I need to reduce the following root to the form a(sqrt)b where 'a' is an integer and 'b' in as small as possible:
    (sqrt)50 = 2(sqrt)25 = 10(sqrt)5 (Is that right?)
    ...
    Hi,

    if I understand your problem right, you want to calculate partially the square-root of 50 (?):

    sqrt(50) = sqrt(25 * 2) = sqrt(25) * sqrt(2) = 5*sqrt(2)

    so, if I understand you right, your result is not right. (so sorry)

    tschüss

    EB
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