Question

The values of $m$ for which both roots of the quadratic equation $x^2-(m-3)x+m=0;m\in R,$ are greater than $2$ is

• A. $[1,9]$
• B. $(7,9]$
• C. $(10,\infty )$
• D. $[9,10)$

Answer 1

The correct option is D $[9,10)$

Given: $x^2-(m-3)x+m=0;m\in R$
On comparing with standard quadratic expression $f\left(x\right)=a{x}^2+bx+c,$ we have $a=1,b=-(m-3),c=m.$

Let, $f\left(x\right)=x^2-(m-3)x+m,$
When both roots of $f\left(x\right)=0$ are greater than $2.$


Condition :
$\left(i\right)D\ge 0$
$\left(ii\right)-\frac{b}{2a}>2$
$\left(iii\right)f\left(2\right)>0$

Now, on solving it,
$\left(i\right)D\ge 0$
$\Rightarrow (m-3{)}^2-4m\ge 0$
$\Rightarrow m^2-6m+9-4m\ge 0$
$\Rightarrow m^2-10m+9\ge 0$
$\Rightarrow (m-1)(m-9)\ge 0$
$\Rightarrow m\in (-\infty ,1]\cup [9,\infty )$

$\left(ii\right)-\frac{b}{2a}>2$
$\Rightarrow \frac{m-3}{2}>2$
$\Rightarrow m-3>4\Rightarrow m>7$
$\Rightarrow m\in (7,\infty )$

$\left(iii\right)f\left(2\right)>0$
$\Rightarrow (2{)}^2-(m-3\left)\right(2)+m>0$
$\Rightarrow 4-2m+6+m>0$
$\Rightarrow -m+10>0\Rightarrow m<10$
$\Rightarrow m\in (-\infty ,10)$

Now, taking intersection of both the condition,
$m\in \left\{\right\{(-\infty ,1]\cup [9,\infty )\}\cap \{(7,\infty )\}\cap \{(-\infty ,10)\left\}\right\}$
$m\in [9,10)$