# Help a Lost Soul

1. Nov 8, 2006

### Erectable

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. Nov 8, 2006

### Sane

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:

\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*}

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
3. Nov 8, 2006

### Werg22

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.

4. Nov 9, 2006

### VietDao29

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? :)