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Thread: Solving for a < 100

  1. Feb 20th, 2003, 03:28 AM #1
    silva
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    Solving for a < 100

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    If a and b, and x are natural numbers, find the possible values of a if a < 100

    How would I solve this? Please help

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  2. Feb 20th, 2003, 11:30 AM #2
    plenderj
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    Well you'd check for which values the square root would give even whole number results.
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  3. Feb 20th, 2003, 12:07 PM #3
    bugzpodder
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    well just say any value of a works.
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  4. Feb 21st, 2003, 02:30 AM #4
    silva
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    1 < a < 99

    A can be from 2 to 98 - I think.

    How can I prove / reason this?

    s.
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  6. Feb 21st, 2003, 05:55 AM #5
    kedaman
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    thats the formula to solve quadratic polynominals
    x^2+bx-a=0
    so in other words
    a=x^2+bx
    put for instance x=1 and for any b in [1,98] you get a in [2,99]
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    reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
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  7. Feb 21st, 2003, 02:03 PM #6
    opus
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    But 1,98 doesn't sound likea natuaral number!
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  8. Feb 21st, 2003, 02:27 PM #7
    kedaman
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    we use commas as decimal delimiter here in Finland, but in math i think its better to stay to . because , is used in other contexts, like [a,b] which means from and including a to and including b
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    writing software in C++ is like driving rivets into steel beam with a toothpick.
    writing haskell makes your life easier:
    reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
    To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
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  9. Feb 21st, 2003, 02:31 PM #8
    opus
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    Sorry
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  11. Feb 22nd, 2003, 10:51 PM #9
    silva
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    OK, I now think 9 < a < 100

    VB Code:
    1. For a = 99 To 1 Step -1

    2.     For b = 11000 To 1 Step -1

    3.     x = ((b + Sqr((b ^ 2) + (4 * a))) / 2)

    4.     If InStr(1, x, ".") = 0 Then List1.AddItem "a: " & a & " b: " & b & " x: " & x

    5.     Next

    6. Next

    How could I explain this or is there a fault in my reasoning?

    s.
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  12. Feb 23rd, 2003, 02:11 PM #10
    prog_tom
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    Originally posted by kedaman
    thats the formula to solve quadratic polynominals
    x^2+bx-a=0
    so in other words
    a=x^2+bx
    put for instance x=1 and for any b in [1,98] you get a in [2,99]
    WRONG, that's not the equation for solving quadratics.

    x = -b +- sqrt(b^2 + 4ac)/2a

    is the Equation for solving quadratics.

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  13. Feb 23rd, 2003, 02:33 PM #11
    kedaman
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    with polynomals ax^2+bx+c=0 yes, but we have the same roots for all n for nx^2+nbx +nc=0.
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    writing software in C++ is like driving rivets into steel beam with a toothpick.
    writing haskell makes your life easier:
    reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
    To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
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  14. Feb 23rd, 2003, 04:10 PM #12
    silva
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    Originally posted by prog_tom
    WRONG, that's not the equation for solving quadratics.

    x = -b +- sqrt(b^2 + 4ac)/2a

    is the Equation for solving quadratics.
    Yes, that's the quadratic formula, but there are other ways in which you can solve quadratics.

    Anyway, I proved it - don't worry

    Thanks for the help.

    s.
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