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Math Help - SAT Question

  1. #1
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    SAT Question

    1. If k and h are constants and x^2 + kx + 7 is equivalent to (x+1)(x+h), what is the value of k?

    (a) 0
    (b) 1
    (c) 7
    (d) 8

    2. If k is a positive integer, which of the following must represent an even integer that is twice the value of an odd integer?
    (a) 2k
    (b) 2k+3
    (c) 2k+4
    (d) 4k+1
    (e) 4k+2

    i thought that the answer was (a) but the actual answer is (e) and i don't know why

    3. If x and y are positive integers and 4(2^x) = 2^y, what is x in terms of y
    (a) y-2
    (b) y-1
    (c) y
    (d) y+1
    (e) y+2

    i didn't know how to solve this because what i did was:
    8^x = 2^ y
    2^3x=2^y
    3x=y
    x=y/3

    but there was no option for that..


    4. If n/(n-1) x 1/n x n/(n+1) = 5/k for positive integers n and k, what is the value of k?
    (a) 1
    (b) 5
    (c) 24
    (d) 25
    (e) 26

    i simplified the whole thing down to:
    k = 5n + 5/n
    and i do not know what to do next
    thank you so much!
    Last edited by juliak; December 28th 2009 at 11:32 PM.
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  2. #2
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    Quote Originally Posted by juliak View Post
    1. If k and h are constants and x^2 + kx + 7 is equivalent to (x+1)(x+h), what is the value of k?

    (a) 0
    (b) 1
    (c) 7
    (d) 8
    Since this is SAT, substitute all the options and get the answer. -1 must be a root of x^2 + kx + 7.
    (a) x^2 + 7, -1 is not a root
    (b) x^2 + x + 7, -1 is not a root
    (c) x^2 + 7x + 7, -1 is not a root
    (d) x^2 + 8x + 7, -1 is a root

    So (d) is the answer.

    Direct method is to compare (x+1)(x+h) with x^2 + kx+7.
    x^2 + (h+1)x + h = x^2 + kx + 7.
    So h+1 = k and h = 7;

    Thus k = 7+1 = 8;

    Quote Originally Posted by juliak View Post
    2. If k is a positive integer, which of the following must represent an even integer that is twice the value of an odd integer?
    (a) 2k
    (b) 2k+3
    (c) 2k+4
    (d) 4k+1
    (e) 4k+2
    i thought that the answer was (a) but the actual answer is (e) and i don't know why
    thank you so much!
    (a) is not the answer. if k is 2, then 2k is not "twice the value of an odd integer"

    However 2k+1 is always odd and therefore 2(2k+1) = 4k+2 is "twice the value of an odd integer". So (e) is correct
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  3. #3
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    Thank you very much but I am a bit confused about the direct method - how did you know that h was 7?
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  4. #4
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    Quote Originally Posted by juliak View Post

    3. If x and y are positive integers and 4(2^x) = 2^y, what is x in terms of y
    (a) y-2
    (b) y-1
    (c) y
    (d) y+1
    (e) y+2

    i didn't know how to solve this because what i did was:
    8^x = 2^ y
    2^3x=2^y
    3x=y
    x=y/3

    but there was no option for that..
    Note that  4(2^x) \ne 8^x

     4(2^x) = 2^2 \cdot 2^x = 2^{x+2}

     2^{x+2} = 2^y

     x + 2 = y

     x = y - 2 Answer A

    4. If n/(n-1) x 1/n x n/(n+1) = 5/k for positive integers n and k, what is the value of k?
    (a) 1
    (b) 5
    (c) 24
    (d) 25
    (e) 26

    i simplified the whole thing down to:
    k = 5n + 5/n
    and i do not know what to do next
    <br />
\frac{\not{n}}{n-1} \cdot \frac{1}{\not{n}} \cdot \frac{n}{n+1} = \frac{5}{k} <br />

    <br />
\frac{n}{(n-1)(n+1)} = \frac{5}{k}

     kn = 5(n^2 - 1)

    5n^2 - kn - 5 = 0

    Your hint here is that k and n are integers. This means the discriminant of this quadratic equation  \Delta = b^2 - 4(a)(c) must be a perfect square. Otherwise n will not be an integer.

    For this equation,
     \Delta = (-k)^2 - 4(5)(-5) \Rightarrow k^2 + 100

    So  k^2 + 100 is a perfect square. Substitute the choices to see which value of k works.

     a) 1^2 + 100 = 101 not a perfect square.
     b) 5^2 + 100 = 125 not a perfect square.
     c) 24^2 + 100 = 676 = 26^2 This is a perfect square. Answer C
     d) 25^2 + 100 = 725 not a perfect square.
     e) 26^2 + 100 = 776 not a perfect square.

    Thank you very much but I am a bit confused about the direct method - how did you know that h was 7?
    Basically you compare the left hand side of the equation with the right hand side.

     x^2 + (h+1)x + h = x^2 + kx + 7
    Comparing  x^2 \Rightarrow x^2 = x^2
    Comparing  x \Rightarrow (h+1)x = (k)x \Rightarrow h+1 = k
    Comparing constants  \Rightarrow h = 7
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