Question

If the roots of $x^2$ - (a - 3)x + a = 0 are such that atleast ont of the roots is greater than 2, then a$ϵ$

• A. [7, 9]
• B. $[7,\infty )$
• C. $[9,\infty )$
• D. (7, 9)

Answer 1

The correct option is C

$[9,\infty )$

$x^2$ - (a - 3)x + a = 0
D= $(a-3{)}^2$ - 4a = $a^2$ - 10a + 9 = (a - 1)(a - 9)
Case 1: Both roots are greater than 2
(i) $D\ge 0\Rightarrow (a-1)(a-9)\ge 0\Rightarrow aϵ(-\infty ,1]\cup [9,\infty )$
(ii) $f\left(2\right)>0\Rightarrow 4-(a-3)2+a>0\Rightarrow a<10$
(iii) $\frac{-B}{2A}>2\Rightarrow \frac{a-3}{2}>2\Rightarrow a>7$
$\therefore aϵ[9,10)$
Case 2 One root is > 2 and other is less than or equal to 2
(i) $\Delta \ge 0\Rightarrow (a-1)(a-9)\ge 0aϵ(-\infty ,1]\cup [9,\infty )$
(ii) $f\left(2\right)\le 0\Rightarrow 4-(a-3)2+a\le 0\to a\ge 10$
$\therefore aϵ[10,\infty )$
So from both the cases we get, $aϵ[9,\infty )$