Discriminat of the Quadratice Formula must be > 1 for this to be true
b^2 - 4ac >0
15^2 - 4(8)(c) >0
225 - 32 c >0
32c <225
c < 225 / 32
c < 225/32 = 7.03125 sooo 7 6 5 4 3 2 1 the product of these is 7!
+14905 Find the product of all positive integer values of $c$ such that $8x^2+15x+c=0$ has two real roots.
Hello dududsu2!
\(8x^2+15x+c=0\)
a b c
\(x =\large {-b \pm \sqrt{b^2-4ac} \over 2a}\)
\(x = {-15 \pm \sqrt{225-32c} \over 16}\)
\( \sqrt{225-32c}\in \mathbb{N}\)
\(c\in \{7\}\)
\(x = \large {-15 \pm \sqrt{225-32\cdot 7} \over 16}\)
\(x\in \{-1,-\frac{7}{8}\}\)
!
