Question

If the roots of $x^2$ - (a - 3)x + a = 0 are such that at least one of the root is greater than 2, then find the range of $a$.

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