How many real roots for $ax^2 + 12x + c = 0$?

Question

If $a$ and $c$ are integers and $2 < a < 8$ and $-1 < c$, how many equations of the form $$ax^2+12x+c=0$$ have real roots?

Answer 1

As @DonAntonio give hint a quadratic eqn have real roots iff $$b^2-4ac\ge0$$ $$144-4ac\ge0\implies ac\le36$$ since $2<a<8$ and $c>-1$ so $a$ will be in set {$3,4,5,6,7$} and $c$ {$0,1,2,3,\cdots$}

from $$ac\le36$$

firstly we will check combination for equality so factors of $36$ in given range $36=3\times 12,4\times 9,6\times 6$

so possible combination of a and c will :

for $a=3,$ $c$ will {$0,1,2,3,4,5,6,7,8,9,10,11,12$}

for $a=4,$ $c$ will {$0,1,2,3,4,5,6,7,8,9$}

for $a=5,$ $c$ will {$0,1,2,3,4,5,6,7$} (this will hold only "<" relation since $5$ is not divider of $36$)

for $a=6,$ $c$ will {$0,1,2,3,4,5,6$}

for $a=7,$ $c$ will {$0,1,2,3,4,5$} (same case as $a=5$)

so possible eqn are $13+10+8+7+6=44$

$44$ equation having a and c in given range have real roots

Answer 2

Hints:

That quadratic's discriminant is

$$\Delta = 144-4ac\ge 144+4a>0\ldots\ldots$$