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Find the product of all positive integer values of c such that 3x^2 + 17x + c = 0 has two real roots.

Apr 6, 2021

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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 $1+2+3+4+5+...+22+23+24=\frac{24(25)}{2}=12*25=300$

Apr 6, 2021