Question

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

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

Answer 1

Answer: (C)

$a\in [9,\infty )$

Given, $x^2-(a-3)x+a=0$

$\Rightarrow D=(a-3{)}^2-4a$

$=a^2-10a+9$

$=(a-1)(a-9)$

Case I:

Both the roots are greater than 2

$D\ge 0,f\left(2\right)>0,-\frac{b}{2a}>2$

$\Rightarrow (a-1)(a-9)\ge 0;4-(a-3)2+a>0;\frac{a-3}{2}>2$

$\Rightarrow a\in (-\infty ,1]\cup [9,\infty );a<10;a>7$

$\Rightarrow a\in [9,10)$ ..........(1)

Case II:

One root is greater than 2 and the other root is less than or. equal to 2. Hence,

$f\left(2\right)\le 0$

$\Rightarrow 4-(a-3)2+a\le 0$

$\Rightarrow a\ge 10$ ..........(2)

From (1) and (2)

$a\in [9,10)\cup [10,\infty )\Rightarrow a\in [9,\infty )$