Find the product of all positive integer values of c such that 3x^2 + 17x + c = 0 has two real roots.
+486 For any quadratic $ax^2+bx+c$, for it to have two real roots, the discriminant, $b^2-4ac$ must be greater than 0. So, we have:
$(17)^2-4(3)(c)>0$
$289-12c>0$
$12c-289<0$
$12c<289$
$c<24$ $\frac{1}{12}$
So the product of all positive integer values of c is $1*2*3*4*5*...*22*23*24$, which is $24!$, a HUGENORMOUS number.
If you meant sum, not product, then it would be $1+2+3+4+5+...+22+23+24=\frac{24(25)}{2}=12*25=300$
