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I Identify same values of two quadratic equation

  1. Jul 6, 2016 #1
    Say there are two quadratic equation
    4x2+560x+296 and
    4x2+4x

    Is there a way to know that two quadratic equation have same values.

    Thank you.
     
  2. jcsd
  3. Jul 6, 2016 #2

    ProfuselyQuarky

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    You mean whether the two expressions are equivalent?
     
  4. Jul 6, 2016 #3
    No. These two equation gives many values starting with x=1 respectively.

    I should have written the second equation with 'y' variable.
     
  5. Jul 6, 2016 #4

    ProfuselyQuarky

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    So ##4x^2+560x+296## and ##4y^2+4y## and you're asking how to find whether ##x=y##?
     
  6. Jul 6, 2016 #5
    If you mean that you want to find which (if any) values satisfy ##4x^2 + 560x + 296 = 4x^2 + 4x##, then any such value must satisfy

    ## 4x^2 + 560x + 296 - 4x^2 - 4x = 556x + 296 = 0##

    The solution is fairly straightforward from there.
     
  7. Jul 6, 2016 #6
    for e.g the value 2600 satisfy both the equation.
    I forgot to mention , only whole number values that are same to both the equation.

    x=4 to 1st equation and x=25 to the 2nd equation results 2600.

    edit: the variable should be a natural number.
     
    Last edited: Jul 6, 2016
  8. Jul 6, 2016 #7

    micromass

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    So you want to find all natural numbers ##(n,m)## such that

    [tex]4n^2 + 560n + 296 = 4m^2 + 4m[/tex]

    Completing the square, we get

    [tex](2n + 140)^2 - 19304 = (2m + 1)^2 - 1[/tex]

    So if you have such numbers ##(m,n)##, then you have also found natural numbers ##(k,l)## (namely ##k = 2n+140## and ##l=2m+1##) such that

    [tex]k^2 - l^2 = 19303[/tex]

    Factorizing into primes, we get

    [tex](k+l)(k-l) = 97\cdot 199[/tex]

    So we have two situations:

    First, we can have ##k-l = 97## and ##k+l = 199##. In this case, we have ##(k,l) = (148,51)##.
    The second solution is where ##k-l = 1## and ##k+l = 19303##. In this case, we have ##(k,l) = (9652, 9651)##.

    Those solutions ##(k,l)## have
    [tex]k=2n+140~\text{and}~l = 2m+1[/tex]
    In the first case, we have ##2n+140 = 148## and ##2m+1 = 51##. This gets us the solutions ##n=4## and ##m=25## you have also found.
    In the second case, we have ##2n + 140 = 9652## and ##2m + 1 = 9651##. This gets us the solutions ##n=4756## and ##m=4852##.

    The previous shows that these are the only solutions.
     
  9. Jul 6, 2016 #8
    Thanks a lot. But it will become impossible to factor larger number (here, it is 19303). There is no formula to factor them.

    So won't be able to tell with equation, which give larger number.
     
  10. Jul 7, 2016 #9

    So the form, for the number ,

    19303 = ( 2 n + 140 )2 - (2m + 1)2.

    This way any larger number can be formed. For primality test, this still, ask the factor of the number.

    For number 104661, I can write

    4 x2 + 1304 x + 1614 = 4 y2 + 4 y

    i.e ( 2 x + 326 )2 - 104662 = ( 2 y +1 )2 - 1

    which will be,

    ( 2 x + 326 ) 2 - ( 2 y + 1 ) 2 = 104661



    If I will be able to solve x and y (both as natural number), I can say 104661 is not a prime number.
     
    Last edited: Jul 7, 2016
  11. Jul 7, 2016 #10
    Also 104661 need to be checked with

    ( 2 x + 325) 2 - ( 2 y )2 = 104661

    as well. These two equation changes gradually with all the numbers.
     
  12. Jul 7, 2016 #11

    Mark44

    Staff: Mentor

    Minor point, but these are NOT equations -- they are algebraic expressions. An equation always has an "equals" symbol in it somewhere. An equation is a statement about two expressions that have the same value.
     
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