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Find the product of all positive integer values of $c$ such that $3x^2+7x+c=0$ has two real roots.

 Jun 11, 2018
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Find the product of all positive integer values of c such that 3x^2+7x+c=0 has two real roots.

 

Hello Lightning!

 

\(3x^2+7x+c=0.\) 

 

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {-7 \pm \sqrt{7^2-4*3c} \over 2*3}\\ x ={ -7 \pm \sqrt{49-12c} \over 6}\)

There are only two real roots if 12c < 49.

\(c \in\{1,2,3,4\}\)

The product is 4! = 24

 

Greetings

laugh  !

 Jun 12, 2018
edited by asinus  Jun 12, 2018

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