Find inequality on a, b and c; if the equation $ax^2+bx+c=0$ has both the roots greater than 2

Question

Question:
If the equation $ax^2+bx+c=0$ has both the roots greater than 2 then,

A. $a(b+c)>0$
B. $a(a+b) > 0$
C. $a(b+c) < 0$
D. $a(a+b)<0$

As, both roots are $> 2$, then they should be greater than $0$ and $1$ too. Also, it follows:

$$af(1) < af(0)$$

(with both sides being positive)

$$a(a+b+c)<ac$$

$$\therefore a(a+b)<0$$

But, it's not the correct option. Where am i wrong?

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The correct option given:

A

Answer 1

If $2<u\le v$ and $f(x)=ax^2+bx+c=a(x-u)(x-v)$

Then, $a^2+a(b+c)= af(1)=a^2(u-1)(v-1)>a^2$

So, $$a(b+c)>0$$ So, A is correct and C is false.

Your reasoning is also valid, so B is false and D is correct.