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Find the product of all positive integer values of $c$ such that $8x^2+15x+c=0$ has two real roots.

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Dec 20, 2020

#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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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

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

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

$$\sqrt{225-32c}\in \mathbb{N}$$

$$c\in \{7\}$$

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

$$x\in \{-1,-\frac{7}{8}\}$$

!

Dec 21, 2020