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Math Help - Golden Ratio equation

  1. #1
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    Golden Ratio equation

    Ive found an equation x-x-1=0 and i need to solve for x to find the golden ratio. I have to find an exact valure for x and a decimal approximation for the golden ratio correct to 6 places. Thankyou
    Also need negative root. Thankyou
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by holmesb View Post
    Ive found an equation x-x-1=0 and i need to solve for x to find the golden ratio. I have to find an exact valure for x and a decimal approximation for the golden ratio correct to 6 places. Thankyou
    Also need negative root. Thankyou
    use the quadratic formula (or completing the square if you prefer)

    By the quadtratic formula:

    x = [1 +/- sqrt(1 + 4)]/2

    = [1 +/- sqrt(5)]/2

    the golden ratio is (1 + sqrt(5))/2

    as a decimal: 1.618034

    for the negative root:

    (1 - sqrt(5))/2 = - 0.618034
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  3. #3
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    When finding the negative root do i use this
    [1 - sqrt(5)]/2
    to find -0.618033988
    correct?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    i recommend you see Golden ratio for more info.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by holmesb View Post
    When finding the negative root do i use this
    [1 - sqrt(5)]/2
    to find -0.618033988
    correct?
    yes, that's what i did
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  6. #6
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    Didnt see that

    Thanks for the help.
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    do you know how to use the quadratic formula and completing the square?
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  8. #8
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    yes for the quadratic equation. Not sure about completing the square.
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  9. #9
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    I also have one more question about fibonacci numbers (didnt know weather to make new thread or post it here):
    I have found this:
    1+1+2=2*3
    1+1+2+3=3*5
    1+1+2+3+5=5*8
    etc
    From this i have to find an equation. This is what i have come up with:
    F(n)*f(n+1)=F(n)
    +f(n-1).........f(1)
    Where i write n+1 i mean the following fiboanci number and n-1 the previous.
    Is this equation somewhat correct if not what changes must be made?
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by holmesb View Post
    I also have one more question about fibonacci numbers (didnt know weather to make new thread or post it here):
    I have found this:
    1+1+2=2*3
    1+1+2+3=3*5
    1+1+2+3+5=5*8
    etc
    From this i have to find an equation. This is what i have come up with:
    F(n)*f(n+1)=F(n)
    +f(n-1).........f(1)
    Where i write n+1 i mean the following fiboanci number and n-1 the previous.
    Is this equation somewhat correct if not what changes must be made?
    the equation seems ok to me.

    i was trying to find a post where someone went through completing the square step by step, but i couldn't find it. i guess you can try online on wikipedia or something if you're interested--you're going to have to learn it eventually

    and yes, your first instinct was right, new questions go in new threads
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  11. #11
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    ok thanks again
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  12. #12
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by holmesb View Post
    Ive found an equation x-x-1=0 and i need to solve for x to find the golden ratio. I have to find an exact valure for x and a decimal approximation for the golden ratio correct to 6 places. Thankyou
    Also need negative root. Thankyou
    i guess i'll try to show you completing the square by doing an example, hopefully you can follow. let's use the characteristic equation for the golden ratio again.

    x^2 - x - 1 = 0

    first thing, we want to make sure the coefficient of x^2 is 1, we're good here for that condition. if we were not, we would divide through by the coefficient to get it to one. so, now on to the method

    x^2 - x = 1 ....................................put the constant on one side
    x^2 - x + (-1/2)^2 = 1 + (-1/2)^2 ....add half the coefficient of x to both sides

    (x - 1/2)^2 = 5/4 ...........................solve the right side and contract the left side by putting x and half the coefficient of x in brackets and squaring them.

    x - 1/2 = +/- sqrt(5/4) ....................take the squareroot of both sides
    x = 1/2 +/- sqrt(5/4) .......................solve for x
    x = 1/2 +/- sqrt(5)/sqrt(4) ................simplify
    x = 1/2 +/- sqrt(5)/2 .......................simplify even more
    x = (1 +/- sqrt(5))/2 ........................combined fractions

    it's always the same steps above, except for the simplification part. if you were doing another problem, you could stop at the "solve for x" line and compute your answers there, but everything before that is routine
    Last edited by Jhevon; March 24th 2007 at 06:47 PM.
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