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Math Help - Another Problem - split from "completing square"

  1. #1
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    Another Problem - split from "completing square"

    alright i have solve it.

    but here is another question which i also don't know how to do.
    can help me.


    Solve the given equations, determine the character of the roots.

    In finding the radius r of a circular arch of height h and span b, an architect used the following formula. solve for h

    <br />
r = \frac {b^2 + 4h^2}{8h}<br />
    Last edited by danielwu; September 8th 2006 at 02:32 AM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by danielwu
    alright i have solve it.

    but here is another question which i also don't know how to do.
    can help me.

    <br />
r = \frac {b^2 + 4h^2}{8h}<br />
    We need a bit of a clue as to what you want done with this.

    RonL
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  3. #3
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    okay i have re-write my post. thanks alot
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  4. #4
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    Quote Originally Posted by danielwu
    alright i have solve it.

    but here is another question which i also don't know how to do.
    can help me.


    Solve the given equations, determine the character of the roots.

    In finding the radius r of a circular arch of height h and span b, an architect used the following formula. solve for h

    <br />
r = \frac {b^2 + 4h^2}{8h}<br />
    Thus,
    8rh=b^2+4h^2
    Thus,
    4h^2-8rh+b^2=0
    The discriminant is,
    \sqrt{64r^2-16b^2}
    Is it positive?
    Are you sure that r>b?
    If thus, then your two roots are real.
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  5. #5
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    Quote Originally Posted by ImPerfectHacker
    Are you sure that r>b?
    If thus, then your two roots are real.
    For a circular arch of radius rspanning b, we must have:

    <br />
b \le 2r<br />

    Its simple geometry.

    RonL
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  6. #6
    Super Member malaygoel's Avatar
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    Quote Originally Posted by CaptainBlack
    For a circular arch of radius rspanning b, we must have:

    <br />
b \le 2r<br />

    Its simple geometry.

    RonL
    what is b here? I don't understand

    Keep Smiling
    Malay
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  7. #7
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    Quote Originally Posted by malaygoel
    what is b here? I don't understand

    Keep Smiling
    Malay
    It is the gap spanned by the arch.

    Think of the arch as forming an arc of a circle, then b is the length of the
    corresponding chord.

    RonL
    Attached Thumbnails Attached Thumbnails Another Problem - split from &quot;completing square&quot;-gash.jpg  
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  8. #8
    Super Member malaygoel's Avatar
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    Quote Originally Posted by CaptainBlack
    It is the gap spanned by the arch.

    Think of the arch as forming an arc of a circle, then b is the length of the
    corresponding chord.

    RonL
    thanks

    Keep Smiling
    Malay
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  9. #9
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    Quote Originally Posted by ThePerfectHacker
    Thus,
    8rh=b^2+4h^2
    Thus,
    4h^2-8rh+b^2=0
    The discriminant is,
    \sqrt{64r^2-16b^2}
    Is it positive?
    Are you sure that r>b?
    If thus, then your two roots are real.


    hi i am not sure if this is the right answer?
    what i have learn in my current chapter for this question is to make use this formula

    <br />
x = -b \pm \sqrt \frac{b^2 - 4ac}{2a}<br />
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  10. #10
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    Quote Originally Posted by danielwu
    hi i am not sure if this is the right answer?
    what i have learn in my current chapter for this question is to make use this formula

    <br />
x = -b \pm \sqrt \frac{b^2 - 4ac}{2a}<br />
    I doubt that, you may have been shown:

    <br />
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}<br />
,

    and the term b^2 - 4ac under the root sign is the
    discriminant that other poster have been referring to.

    RonL
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  11. #11
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    Quote Originally Posted by danielwu
    hi i am not sure if this is the right answer?
    what i have learn in my current chapter for this question is to make use this formula

    <br />
x = -b \pm \sqrt \frac{b^2 - 4ac}{2a}<br />
    You mean,
    x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}
    ---
    That is what I did.
    You only look at the expression under the radical.
    b^2-4ac
    This tells you its signs.
    If it is negative then you have imaginaries.
    If it is zero then you have one a singal solution to quadradic.
    It it is positive then you have two distint solution.
    And if the positive term is a perfect square then you have rational solutions.
    To summerize,
    b^2-4ac\left\{ \begin{array}{cc}<0&\mbox{ imaginary}\\ =0&\mbox{a single rational solution}\\ >0&\left\{ \begin{array}{cc}\mbox{square}&\mbox{rational}\\ \mbox{not a square}&\mbox{irrational} \end{array} \end{array} \right
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  12. #12
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    Quote Originally Posted by ThePerfectHacker
    Thus,
    8rh=b^2+4h^2
    Thus,
    4h^2-8rh+b^2=0
    The discriminant is,
    \sqrt{64r^2-16b^2}
    Is it positive?
    Are you sure that r>b?
    If thus, then your two roots are real.


    hi i am not sure if this is the right answer?
    what i have learn in my current chapter for this question is to make use this formula

    <br />
x = -b \pm \sqrt \frac{b^2 - 4ac}{2a}<br />

    Example:

    If a = 3, b= -3, c = -2 for  3y^2 - 3y -2 = 0

    Then what is for

    r = \frac {b^2 + 4h^2}{8h}<br />
    a = ?, b = ?, c = ?
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  13. #13
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker

    ..The discriminant is,
    \sqrt{64r^2-16b^2}
    The discriminant is {64r^2-16b^2}

    RonL
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  14. #14
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    Quote Originally Posted by ThePerfectHacker
    Thus,
    8rh=b^2+4h^2
    Thus,
    4h^2-8rh+b^2=0
    The discriminant is,
    \sqrt{64r^2-16b^2}
    Is it positive?
    Are you sure that r>b?
    If thus, then your two roots are real.


    hi i am not sure if this is the right answer?
    what i have learn in my current chapter for this question is to make use this formula

    <br />
x = -b \pm \sqrt \frac{b^2 - 4ac}{2a}<br />

    Example:

    If a = 3, b= -3, c = -2 for
     3y^2 - 3y -2 = 0

    Then what is for

    r = \frac {b^2 + 4h^2}{8h}<br />
    a = ?, b = ?, c = ?
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  15. #15
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    Quote Originally Posted by danielwu

    Then what is for

    r = \frac {b^2 + 4h^2}{8h}<br />
    a = ?, b = ?, c = ?
    Multiply by 8h to get,
    8hr=b^2+4h^2
    Thus,
    4h^2-8hr+b^2=0
    Thus,
    a=5,b=-8h,c=b^2
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