Question

If the quadratic equation $a{x}^2-bx+12=0$, where $a$ and $b$ are positive integers not exceding 10 , has roots both greater than 2 , then the number of possible ordered pair $(a,b)$ is

• A. 1
• B. 2
• C. 4
• D. 8

Answer 1

Answer: (A)

$\Theta$ Both roots are greater than 2
$\therefore$
(i) a f (2) >0
$\Rightarrow a(4a-2b+12)>0\Rightarrow 2a-b+6>0$
(ii) $D\ge 0\Rightarrow b^2-48a\ge 0\Rightarrow b\ge 4\sqrt{3}a$
(iii) $\frac{-(-b)}{2a}>2\Rightarrow b>4a$
$\therefore b\ge 4\sqrt{3}a$
If $a=1,b\ge 4\sqrt{3},b\ge 7$
$\Theta 2a-b+6>0$
$\therefore a=1,b=7$ satisfies it
$\therefore$ Number of ordered pairs $(a,b)=1$