If both the roots of $x2-ax+a=0$ are greater than 2 then determine the interval in which $a$ lies.
I used the following two conditions:
$\alpha\beta>2$ and $D\ge0$
Thus I obtained $a>2$ and $a \in (-infty,0)\cup(4,infty)$ and their intersection gives: $(4,\infty)$ but answer shocked me. It says: "No such a exists"

If both the roots of $x^2-ax+a=0$ are greater than 2 then determine the interval in which $a$ lies.
I used the following two conditions:
$\alpha\beta>2$ and $D\ge0$
Thus I obtained $a>2$ and $a \in (-\infty,0)\cup(4,\infty)$ and their intersection gives: $(4,\infty)$ but answer shocked me. It says: "No such a exists"

The equation $ax^2+bx+c=0$ has real and positive roots. Prove that the roots of the equation $ax^2+a(3b-2c)x+(2b-c)(b-c)+ac=0$ are real and positive.
Now we have,
$b^2-4ac>0$.
$(-b/a)>0$
$c/a>0$
I need to prove :
$a^2(9b^2+9c^2-18bc)-4a^2(2b^2-3bc+c^2+ac)>0$

I managed to prove all the terms positive (you may ask me if you suspect) but am stuck on the last term $-4a^3c$. How do i prove that $-4a^3c$, too, is positive?

The equation $ax^2+bx+c=0$ has real and positive roots. Prove that the roots of the equation $a^2x^2+a(3b-2c)x+(2b-c)(b-c)+ac=0$ are real and positive.
Now we have,
$b^2-4ac>0$.
$(-b/a)>0$
$c/a>0$
I need to prove :
$a^2(9b^2+9c^2-18bc)-4a^2(2b^2-3bc+c^2+ac)>0$

I managed to prove all the terms positive (you may ask me if you suspect) but am stuck on the last term $-4a^3c$. How do i prove that $-4a^3c$, too, is positive?