Question

If fx = ax 2 + bx + c has no real roots and a + b + c < 0 then prove c <0

Answer 1

given equation is: 
f(x)=ax^2+bx+c
Now that it has no real roots, 
discriminant<0

therefore- b^2-4ac<0
b^2<4ac
|b|<2√ac
|b|<a+c
note that: for any two no. x nd y we hv x+y>2√xy
so, b<a +c, and
-b<a+c
but
-b<a+c, so,
a+b+c>0
b<a+c
c>b-a
therefore, value of c must be greater than b-a