Expert: Paul Klarreich - 9/19/2007

Question
Hi Paul,

My Q is :

Let 84x^2 +kx – 165 =0 be a quadratic eqn. Find the number of possible values of “k” for which the roots of the quadratic eqn are RATIONAL numbers, where k is any integer

Approach :
I think the b^2 - (4*a*c) shuld be a perfact squere.... But then cant proceed..

Answer
Questioner:   pnd
Category:  Advanced Math
Private:  No
 
Subject:  roots and equation
Question:  Hi Paul,

My Q is :

Let 84x^2 +kx - 165 =0 be a quadratic eqn. Find the number of possible values of k for which the roots of the quadratic eqn are RATIONAL numbers, where k is any integer

Approach :
I think the b^2 - (4*a*c) shuld be a perfact square.... But then can't proceed..
....................................
Hi, pnd,

First, a little note. If you use special characters, like the ones you get out of the Character Map feature of Windows, they get messed up.  Stick with the keyboard until this site stops mangling those things.  

In your equation,

84x^2 + kx - 165 =0

you are indeed correct -- the discriminant, b^2 - 4ac, must be a perfect square.  Here it is :

k^2 - 4(84)(-165) =

k^2 + 4(84)(165) =

k^2 + 4(4)(21)(5)(33) =

k^2 + 4(4)(3)(7)(5)(3)(11) =

k^2 + 4^2 3^2 7 5 11 = A^2

However, I don't think this is going to lead to anything useful.  There might be a fancy (and deep and difficult) number-theoretical result that works here, but I don't know one.
..............................

Instead, suppose we try to factor  84x^2 + kx - 165.  If you had your value of k, you would proceed something like this:

The prime factorization of 84 =  2^2 3 7
The divisors of 84 are  1,2,3,4,6,7,12,21,28,42,84

The possible pairs of factors of 84 are:

1*84,  2*42, 3*28, 4*21, 7*12 -- six in all. Each one, such as 3*28, corresponds to writing a pair of factors of your polynomial such as:

(3x ....)(28x ....)

Now how about  165?

The prime factorization of 165 = 3 5 11
The divisors of 165 are  1,3,5,11,15,33,55,165

The possible pairs of factors of 165 are:

1*165,  3*55,  5*33, 11*15 -- four in all, EIGHT if you recall that the product is really  -165, so we could have, for example, both   -5*33  and  -33*5.

Each one of THESE can be paired with one of the six pairs of factors of 84, for example:

(3x - 5)(28x + 33)  or (3x + 5)(28x - 33)

and if you multiply any one of these out, you get your middle term, and you get your value of k.

So I believe that, with six ways to factor the first term and eight ways to factor the last term, there are 48 possible factorizations of the polynomial and therefore 48 possible values of k.  (You did say 'k is an integer', so it can be negative.)