$c>0$ and $3ax^2+4bx+c =0$ has no real roots, then?

Question

If $c>0$ and the equation $3ax^2+4bx+c =0$ has no real roots, then:

A)$ 2a+c>b $

B )$a+2c>b$

C) $3a+c>4b$

D)$a+3c<b$

My try

$b^2-4ac <0 \to 16b^2-12ac <0 $

I.e $4b^2 <3ac $

But the options are different from my result

What is wrong with my approach?

Answer 1

Since the options are linear in the coefficients, a sensible guess is that it wants you to put a particular value of $x$ in.

Since $c>0$ and $3ax^2+4bx+c=0$ has no real roots, $3ax^2+4bx+c$ must be positive for any real $x$ (it is positive when $x=0$, and then if it were negative somewhere, since it is continuous it would have to be zero somewhere). I.e., for any real $x$, $$ 3ax^2+4bx+c > 0 $$ Putting $x=-1$ gives C).

Answer 2

It is not in contradiction with what you obtained :

Using the AM-GM,

$$ 3a +c \geq 2 \sqrt{3ac} > 2\sqrt{4b^2} = 4b $$

Then, C is correct!