+0  
 
 #1
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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!

 Dec 21, 2020
edited by Guest  Dec 21, 2020
 #2
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+1

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

\(x =\large {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(x = {-15 \pm \sqrt{225-32c} \over 16}\)

\( \sqrt{225-32c}\in \mathbb{N}\)

\(c\in \{7\}\)

\(x = \large {-15 \pm \sqrt{225-32\cdot 7} \over 16}\)

\(x\in \{-1,-\frac{7}{8}\}\)

laugh  !

 Dec 21, 2020

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