Let p and q be the two distinct solutions to the equation \((x-3)(x+3)=21x-63\). If \(p>q\), what is the value of \(p-q\)?
Hint: Does not equal 7
The roots are 3 and 5, so p - q = 5 - 3 = 2.
That's incorrect, sorry; can someone else pls help
The quadratic arranges to \(x^2 - 15x + 36 = 0\)
Then by the quadratic formula,
\(x = \frac{15 \pm \sqrt{15^2 - 4 \cdot 36}}{2} = \frac{15 \pm 9}{2} = 12, 3\)
Therefore, \(p - q = 12 - 3 = \boxed{9}\)
also incorrect
(x - 3)(x + 3) = 21x - 63
x2 - 9 = 21x - 63
x2 - 21x + 54 = 0
p = [ 21 + sqrt( 212 - 4·1·54 ) ] / 2 = [21 + sqrt( 225 ) ] / 2 = [ 21 + 15 ] / 2 = 18
q = [ 21 - sqrt( 212 - 4·1·54 ) ] / 2 = [21 - sqrt( 225 ) ] / 2 = [ 21 - 15 ] / 2 = 3
p - q = 18 - 3 = 15
+660 
