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Math Help - please help me with these problems.

  1. #1
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    Exclamation please help me with these problems.

    please help me with the following problems:
    1.For x2- 25 = 0, x=

    2.Add and simplify: 5/6+1/4

    3.Simplify the square root of 27 + the square root of 75.

    4.solve:3xsquared-2x-2=0

    5.Find N for log2N=3

    6.Find N for logb32=5

    thanks so much for your help!!!
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  2. #2
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    Quote Originally Posted by yellowjello View Post
    please help me with the following problems:
    1.For x2- 25 = 0, x=

    2.Add and simplify: 5/6+1/4

    3.Simplify the square root of 27 + the square root of 75.

    4.solve:3xsquared-2x-2=0

    5.Find N for log2N=3

    6.Find N for logb32=5

    thanks so much for your help!!!
    1) Factor. Difference of squares.
    2) Common denominator.
    3) Common factor. sqrt(27) = sqrt(3*9) = sqrt(3)*sqrt(9) ... Same thing with sqrt(75)
    4) Factor or quadratic formula
    5) Definition of logarithms.
    6) Where's the N?
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  3. #3
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    answer to #3

    okay so i'm trying to factor #3
    and i'm getting as far as (3x-2) (x-1) I know the signs are probably messed up and the 1 &2 are messed up, I know that the #'s that go in the spots of the 1 & 2 are supposed to multiply to equal -2 and add to equal the other -2 I just don't knwo how to get there. help?
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  4. #4
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    quad form

    yes i know the quad form where do i plug everything in? that i do not remember
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by yellowjello View Post
    yes i know the quad form where do i plug everything in? that i do not remember
    if you knew the formula, you would know where to plug everything in

    the formula says the solution(s) for ax^2 + bx + c = 0 is/are given by

    x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}

    now, what is a, b and c in your problem? then what would x be?
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  6. #6
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    okay

    yeah nevermind i just tried it, is it no solution?
    that's what i'm getting...
    as long as this is right that's the answer i'm getting
    a=3
    b=2
    c=-2
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by yellowjello View Post
    yeah nevermind i just tried it, is it no solution?
    that's what i'm getting...
    as long as this is right that's the answer i'm getting
    a=3
    b=2
    c=-2
    according to what you typed originally, b = -2. solutions exist in that case
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  8. #8
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    I had b=-2...



    so the top is set up as -2 plus or minus the sqrt of 2squared -4 (3)(-2)
    and the bottom as 2(3)
    is this right?
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by yellowjello View Post
    I had b=-2...



    so the top is set up as -2 plus or minus the sqrt of 2squared -4 (3)(-2)
    and the bottom as 2(3)
    is this right?
    yes. and that has 2 solutions.
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