# finding integers

• Sep 20th 2008, 11:53 AM
pop_91
finding integers
(px + 2)2 = 9x2 +qx + 4

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

thanks for any help
• Sep 20th 2008, 11:59 AM
Jhevon
Quote:

Originally Posted by pop_91
(px + 2)2 = 9x2 +qx + 4

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

thanks for any help

expand the left hand side and compare coefficients

an alternate way would be to complete the square of the right hand side, and then you will clearly see what p and q are
• Sep 20th 2008, 12:00 PM
Moo
Hello,
Quote:

Originally Posted by pop_91
(px + 2)2 = 9x2 +qx + 4

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

thanks for any help

Is it $(px+2)^2=9x^2+qx+4$ ?

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+12x+4.

And 3x+2=px+2
• Sep 20th 2008, 12:02 PM
Matt Westwood
$(px + 2)^2 = 9x^2 +qx + 4$

First you can multiply out and simplify:

$p^2x^2 + 4px + 4 = 9x^2 +qx + 4$

$(p^2 - 9)x^2 + (4p - q)x = 0$

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 $x^2$: $p^2 - 9 = 0$

Coefficients of $x$: $4p - q = 0$

take it away, maestro ...
• Sep 20th 2008, 12:03 PM
icemanfan
Is the 9x2 $9 \cdot 2$ or $9x \cdot 2$ ?

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

$2(px + 2) = 18x + qx + 4$

$2px + 4 = 18x + qx + 4$

$2px = 18x + qx$

$2px - qx = 18x$

$x(2p - q) = 18x$

$2p - q = 18$

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 PM
pop_91
supposed to be
(px + 2) squared = 9x squared + qx + 4