(px + 2)2 = 9x2 +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
Hello,
Is it $\displaystyle (px+2)^2=9x^2+qx+4$ ?
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
$\displaystyle (px + 2)^2 = 9x^2 +qx + 4$
First you can multiply out and simplify:
$\displaystyle p^2x^2 + 4px + 4 = 9x^2 +qx + 4$
$\displaystyle (p^2 - 9)x^2 + (4p - q)x = 0$
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 $\displaystyle x^2$: $\displaystyle p^2 - 9 = 0$
Coefficients of $\displaystyle x$: $\displaystyle 4p - q = 0$
take it away, maestro ...
Is the 9x2 $\displaystyle 9 \cdot 2$ or $\displaystyle 9x \cdot 2$ ?
I'm going to assume it's the latter, so you have
$\displaystyle 2(px + 2) = 18x + qx + 4$
$\displaystyle 2px + 4 = 18x + qx + 4$
$\displaystyle 2px = 18x + qx$
$\displaystyle 2px - qx = 18x$
$\displaystyle x(2p - q) = 18x$
$\displaystyle 2p - q = 18$
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).