Question

The trinomial $a{x}^2+bx+c$ has no real roots, $a+b+c<0.$ Find the sign of the number $c$.

Answer 1

The quadratic equation $a{x}^2+bx+c$ has no real roots and $a+b+c<0.$

If it has no real roots, then $D<0$

$b^2-4ac<0$

$b^2<4ac$

So the sign of $a$ and $c$ are the same, either both positive or both negative.

If both are positive then $0<a+c<-b$ because $a+b+c<0$

$0<(a+c{)}^2<b^2<4ac$

$(a+c{)}^2<4ac$

$a^2+2ac+c^2<4ac$

$a^2-2ac+c^2<0$

$(a-c{)}^2<0$

This is clearly not possible, so $a$ and $c$ can not be both positive.

So the sign of $c$ is negative.