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Math Help - Square root help...

  1. #1
    Junior Member alwaysalillost's Avatar
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    Post Square root help...

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    Can you please help me find the square root to these problems? Thanks so much!
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  2. #2
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    Quote Originally Posted by alwaysalillost View Post
    1.

    2.

    Can you please help me find the square root to these problems? Thanks so much!
    Hello,

    are you supposed to expand the brackets?

    If so:

    to #1: The result is a diffrence of squares:

    (√(2) - 3)*(√(2) + 3) = (√(2)) - 3 = 7

    to #2: This is a complete square. Use the binomal formula:

    (√(3) - √(5))*(√(3) - √(5)) = (√(3)) - 2*√3 * √5 + (√(5)) = 8 - 2*√(15)
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  3. #3
    Junior Member alwaysalillost's Avatar
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    Quote Originally Posted by earboth View Post
    to #1: The result is a diffrence of squares:

    (√(2) - 3)*(√(2) + 3) = (√(2)) - 3 = 7

    If you can please explain to me how you got this answer I'd be very grateful. Thanks in advance!
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by alwaysalillost View Post
    If you can please explain to me how you got this answer I'd be very grateful. Thanks in advance!
    do you know what "the difference of two squares" refers to?

    as it's name suggests, we have the difference of two squares when we subtract one squared number from another. now it turns out, that this operation has a pretty neat factorization. in general:

    x^2 - y^2 = (x + y)(x - y)

    meaning, if you expand the brackets on the right, you get the thing on the left. (do you know how to expand brackets? if that's your problem, you should say so). so all earboth did, was to work this equation backwards. that is, he went from the factorized version to the expanded version:

    (x + y)(x - y) = x^2 - y^2

    in this particular question, your x was √(2) and your y was 3, so all he had to do was to find what (√(2)) was and what 3 was and subtract the latter from the first.

    know, when we square a number that is square-rooted, we simply remove the square root, why is that? well, i believe you asked a question about going from radicals to exponents a while back (see http://www.mathhelpforum.com/math-help/pre-algebra-algebra/14933-exponential-form.html#post51855 ) so you should recall, that we can think of a squareroot of a number as that number to the 1/2 power. now if we raise a number to a power, we multiply the power, so we multiply the 1/2 by 2 and get 1, so we end up with the number to the 1 power, which is just the number. that is:

    (√(2)) = [(2)^(1/2)]^2 = 2^[(1/2) * 2] = 2^1 = 2

    and of course you know that 3^2 = 9

    so we get:

    (..x..-.y).*.(..x..+..y.).=.....x...-..y
    (√(2) - 3)*(√(2) + 3) = (√(2)) - 3 = 2 - 9 = -7


    Challenge:

    what do you think (cuberoot(5))^3 = ?

    what do you think (sqrt(3))^3 = ? as a radical? as a number raised to an exponent?
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  5. #5
    Junior Member alwaysalillost's Avatar
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    Thanks a lot Jhevon!
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