(px+ 2)2 = 9x2 +qx+ 4

Wherepandqare positive integers find the values ofpandq

not quite sure how to go about this one

thanks for any help

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- Sep 20th 2008, 11:53 AMpop_91finding integers
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*px*+ 2)2 = 9*x*2 +*qx*+ 4

Where*p*and*q*are positive integers find the values of*p*and*q*

not quite sure how to go about this one

thanks for any help - Sep 20th 2008, 11:59 AMJhevon
- Sep 20th 2008, 12:00 PMMoo
Hello,

Is it ?

The rough way is to develop the LHS and recognize the coefficients.

Another way is to find q such that 9x²+qx+4 is a perfect square, since the LHS is a perfect square.

Note that it is equal to (3x)²+qx+2²

So it would be (3x+2)²=9x²+**12**x+4.

And 3x+2=px+2 - Sep 20th 2008, 12:02 PMMatt Westwood

First you can multiply out and simplify:

As this is supposed to be an identity for all x, you can equate powers of x on both sides. As the RHS is zero (because everything's been gathered to the left), you can say:

Coefficients of :

Coefficients of :

take it away, maestro ... - Sep 20th 2008, 12:03 PMicemanfan
Is the 9x2 or ?

I'm going to assume it's the latter, so you have

So basically any positive integers for p and q such that the above equation is true will suffice. (p, q) = (10, 2) is one such solution (and provides the lowest possible values for p and q). - Sep 20th 2008, 12:09 PMpop_91
supposed to be

(px + 2) squared = 9x squared + qx + 4