Question

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

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

Answer 1

The correct option is A

$[9,\infty )$

Given: $x^2-(a-3)x+a=0$
Now, discriminant of the equation is given as:
$D=(a-3{)}^2-4a=a^2-10a+9=(a-1\left)\right(a-9)$

Case 1: Both roots are greater than $2$

$\left(i\right)D\ge 0\Rightarrow (a-1)(a-9)\ge 0$
$\Rightarrow a\in (-\infty ,1]\cup [9,\infty )$

$\left(ii\right)f\left(2\right)>0\Rightarrow 4-(a-3)2+a>0$
$\Rightarrow a<10$

$\left(iii\right)-\frac{b}{2a}>2\Rightarrow \frac{a-3}{2}>2$
$\Rightarrow a>7$
$\therefore a\in [9,10)$

Case 2 :- One root is greater than $2$ and other is less than or equal to $2$

$\left(i\right)D\ge 0\Rightarrow (a-1)(a-9)\ge 0a\in (-\infty ,1]\cup [9,\infty )$

$\left(ii\right)f\left(2\right)\le 0\Rightarrow 4-(a-3)2+a\le 0$
$\Rightarrow a\ge 10$
$\therefore a\in [10,\infty )$

Thus, from the intersection of both the cases, we get $a\in [9,\infty ).$