1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help a Lost Soul

  1. Nov 8, 2006 #1
    The problem is:

    S = an^2 + bn

    Where a and b are constants.

    Possible values for S are: 6, 15, 27, 42 and 60
    Possible values for n are: 1 when S=6, 2 when S=15, 3 when S=27, 4 when S=42 and 5, when S=60.

    I am asked to find the values of a and b.

    The way I tried to tackle it was sort it out into quadratic form:

    an^2 + bn - s = 0

    I then substituted suitable values for n and s. After that, I rearrange the equation so I can get the value of of either b or a and try to solve it using simultaneous equations.

    I've tried many times to find the values of a or b but everytime I end up with a 0 = 0 scenario and in the case of S=60 and n=5:

    0(5^2) + 0(5) - 60 does not equal 0.

    My other idea is to use quadratic formula somehow to find the coefficients of n^2 and n but I can't find anything about solving it this way.

    I would appreciate it if somebody with a good sturdy brain could please help a dumb richard like myself.
     
  2. jcsd
  3. Nov 8, 2006 #2
    I'm afraid if I tell you, you're going to murder yourself. You've definitely overcomplicated the entire problem.

    Given that information we can sub in values for S and n respectively to develop the following equations:

    [tex]\begin{align*}
    6 &= a(1)^{2} + b(1)\\
    &= a + b\\
    \\
    15 &= a(2)^{2} + b(2)\\
    &= 4a + 2b\\
    \\
    27 &= a(3)^{2} + b(3)\\
    &= 9a + 3b\\
    \\
    42 &= a(4)^{2} + b(4)\\
    &= 16a + 4b\\
    \\
    60 &= a(5)^{2} + b(5)\\
    &= 25a + 5b\\
    \end{align*}[/tex]

    We don't even need all of those equations. Three fifths of those equations are redundant. We can take any two (it's easiest to take the first two) and solve for a and b with the classic "system of equations". Elimination, to be precise. I hope this is all you need. You should be able to take it from here.
     
    Last edited: Nov 8, 2006
  4. Nov 8, 2006 #3
    This problem ask you to use two equations to solve two unknowns, or else there's an infinite number of solutions for a and b.
     
  5. Nov 9, 2006 #4

    VietDao29

    User Avatar
    Homework Helper

    Ok, as others have pointed out. So just choose 2 equations randomly, the 2 that you think you like best from the 5 equations above. Then solve for a, and b.
    Then we use the 3 rest equations to test, i.e, we'll plug the value of a, and b in the 3 equations. If they all hold, then a, and b are your solutions. If one of them does not hold, then this system of equations has no solution. Can you get this?
    Canyou go from here? :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Help a Lost Soul
  1. Help ! (Replies: 6)

  2. How many souls? (Replies: 9)

Loading...