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Math Help - factor

  1. #1
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    factor

    hello all, i was scanning my memory for some of the questions that were on my math exam, could you please confirm the following answers:

    1.) is k^2 - 2k + 4 a factor of K^3 - 8? My answer: NO.

    2.) For a solution to have two real and unequal solution, the discriminant should be? my answer: GREATER THAN 0

    3.) Simplify (a-b)^2 - (a+b)^2. My answer: -4ab

    4.) There are two decks of 52 playing cards, find the probability of obtaining a club or a king: my answer: 16/52 or 4/13.

    5.) x^2 - 6x + 1 in turning point form has values of a,h and k as? my answer: a = 1, h = 3, k = 2.

    6.) x^2 - 5x +2 factorised completely is? My answer: (x-5+root7/2) (x-5-root7/2)

    thanks guys, i hope i did well
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by andrew2322 View Post
    hello all, i was scanning my memory for some of the questions that were on my math exam, could you please confirm the following answers:

    1.) is k^2 - 2k + 4 a factor of K^3 - 8? My answer: NO.

    2.) For a solution to have two real and unequal solution, the discriminant should be? my answer: GREATER THAN 0

    3.) Simplify (a-b)^2 - (a+b)^2. My answer: -4ab

    4.) There are two decks of 52 playing cards, find the probability of obtaining a club or a king: my answer: 16/52 or 4/13.


    5.) x^2 - 6x + 1 in turning point form has values of a,h and k as? my answer: a = 1, h = 3, k = 2.
    your k is wrong

    6.) x^2 - 5x +2 factorised completely is? My answer: (x-5+root7/2) (x-5-root7/2)

    thanks guys, i hope i did well
    nope
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  3. #3
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    hey

    how is question 5 wrong?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by andrew2322 View Post
    how is question 5 wrong?
    x^2 - 6x + 1 = (x - 3)^2 - 8
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  5. #5
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    Jhevon pointed out your errors (and my work agrees), but I'll see if I can lead you on the right track:

    Quote Originally Posted by andrew2322 View Post
    5.) x^2 - 6x + 1 in turning point form has values of a,h and k as? my answer: a = 1, h = 3, k = 2.
    The standard form is y = a(x - h)^2 + k. To get the original equation in this form, we must complete the square. Ignore the constant term for now, and concentrate on the expression x^2 - 6x. We want to add a term to this that will give us a perfect square. If we consider expressions of the form x^2 + bx, then the constant term that we need to add is \left(\frac b2\right)^2. Why? Because a perfect square of the form \left(m + n\right)^2 = m^2 + 2mn + n^2, which gives m^2 = x^2,\;2mn = bx,\text{ and so }n^2 = \left(\frac{bx}{2m}\right)^2 = \left(\frac{bx}{2x}\right)^2 = \left(\frac b2\right)^2.

    So, we add \left(\frac b2\right)^2 to our expression to form a perfect square, and then we can factor:

    x^2 - 6x = x^2 - 6x + \left(\frac62\right)^2 - \left(\frac62\right)^2

    = x^2 - 6x + 9 - 9

    = x^2 - 2\cdot3\cdot x + 3\cdot3 - 9

    = (x - 3)^2 - 9

    So going back to our original equation, we find that

    x^2 - 6x  + 1 = (x - 3)^2 - 9 + 1 = (x - 3)^2 - 8

    and \left\{\begin{array}{rcr}<br />
a & = & 1\\<br />
h & = & 3\\<br />
k & = & -8<br />
\end{array}\right..

    Quote Originally Posted by andrew2322 View Post
    6.) x^2 - 5x +2 factorised completely is? My answer: (x-5+root7/2) (x-5-root7/2)
    This one is, in form, close enough to the actual solution that I might suspect a typo. If not, just go over your work once more as you probably just made a silly mistake somewhere.
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