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Math Help - Quadratic formulas

  1. #1
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    Quadratic formulas

    Would like a bit of help on a maths quiz. No 1...I know what I am supposed to do, factorising, and putting a,b,c into the x = b+/- sqrt(b^2 etc etc formula... but im getting the wrong answer. So maybe my factorising is wrong... Can someone show me a step by step process for this question.

    And No 2, Im a bit lost, so any help would be appreciated.



    1)Give the turning point of the following quadratic and state whether it is a maximum or minimum.

    y = (−8+9x) (−4x−5)


    2) If a^2 = b^2 + 3 and a − b = 6, what is a + b? Please give an exact answer, using fractions if necessary; no decimals.
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  2. #2
    Member SENTINEL4's Avatar
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    Quote Originally Posted by Maccabhoy View Post
    2) If a^2 = b^2 + 3 and a − b = 6, what is a + b? Please give an exact answer, using fractions if necessary; no decimals.
    a-b=6 \Rightarrow a=b+6 put this to a^2=b^2+3 and you have

    (b+6)^2=b^2+3 \Rightarrow b^2+12b+36=b^2+3 \Rightarrow 12b+36=3 \Rightarrow 12b=3-36 \Rightarrow 12b=-33 \Rightarrow b=\frac{-33}{12}

    Now replace b=\frac{-33}{12} back to a=b+6 and you find a

    a=\frac{-33}{12}+6 \Rightarrow a=\frac{-33}{12}+\frac{72}{12} \Rightarrow a=\frac{39}{12}

    So now

    a+b=\frac{39}{12}+(\frac{-33}{12})=\frac{39}{12}-\frac{33}{12}=\frac{6}{12}=\frac{1}{2}
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  3. #3
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    Here's another way to do #2:

    If a^{2} = b^{2} + 3
    then a^{2} - b^{2} = 3.
    LHS is a difference of 2 squares. Factor into (a - b)(a + b) = 3.
    Replace (a - b) with 6: 6(a + b) = 3.
    Divide both sides by 6 to get a + b = \frac{1}{2}.


    01
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  4. #4
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    Thanks guys. Much appreciated.

    Any ideas on No 1?
    Last edited by mr fantastic; May 13th 2009 at 06:06 AM. Reason: Merged posts
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  5. #5
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    Just a couple more...

    Give the turning point of the following quadratic and state whether it is a maximum or minimum.
    1) y = −4 x^2−9 x−7

    and Find all solutions of the equation:
    2) x^3−5 X^2 +4 x=0
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  6. #6
    Member SENTINEL4's Avatar
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    Quote Originally Posted by Maccabhoy View Post
    and Find all solutions of the equation:
    2) x^3−5 X^2 +4 x=0
    x^3-5x^2+4x=0 \Rightarrow x(x^2-5x+4)=0
    Now it would be x=0 or x^2-5x+4=0

    By solving the x^2-5x+4=0 ypu find x=1 , x=4

    So the solutions are : x=0 , x=1 , x=4
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  7. #7
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    Quote Originally Posted by SENTINEL4 View Post
    x^3-5x^2+4x=0 \Rightarrow x(x^2-5x+4)=0
    Now it would be x=0 or x^2-5x+4=0

    By solving the x^2-5x+4=0 ypu find x=1 , x=4

    So the solutions are : x=0 , x=1 , x=4
    Thanks for that!

    Would you have an idea about my original No 1 question...


    1)Give the turning point of the following quadratic and state whether it is a maximum or minimum.

    y = (−8+9x) (−4x−5)

    I was solving it by factorisation, but am getting a wrong answer...

    Is this wrong... y = -36x^2 - 12x + 40.... and then working from there..
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  8. #8
    Member SENTINEL4's Avatar
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    Quote Originally Posted by Maccabhoy View Post

    1)Give the turning point of the following quadratic and state whether it is a maximum or minimum.

    y = (−8+9x) (−4x−5)
    Well i don't remember well, but i think that turning point has something to do with the second derivative.
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  9. #9
    Member TheMasterMind's Avatar
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    Quote Originally Posted by SENTINEL4 View Post
    Well i don't remember well, but i think that turning point has something to do with the second derivative.
    y = (−8+9x) (−4x−5)

    simple foil gives:

    y=-36x^2-13x+40

    complete the square, this will give you the quadratic in vertex form, which is the turning point.

    although after the foil, it is evident that it is a negative x value, this a maximum because the parabola is opening down
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