Question

Find the value of $m$ such that one root is smaller than $2$ and other root is greater than $2$ for the quadratic equation $x^2-(m-3)x+m=0$ $(m\in R)$

• A. $(9,10)$
• B. $(-\infty ,10)$
• C. $(1,10)$
• D. $(10,\infty )$

Answer 1

The correct option is D $(10,\infty )$

Let, $f\left(x\right)=x^2-(m-3)x+m$


Condition :

$\left(i\right)D>0$
$\left(ii\right)f\left(2\right)<0$

Solving these conditions, we get:
$\left(i\right)D>0$
$\Rightarrow (m-3{)}^2-4m>0$
$\Rightarrow m^2-6m+9-4m>0$
$\Rightarrow m^2-10m+9>0$
$\Rightarrow (m-1)(m-9)>0$
$m\in (-\infty ,1)\cup (9,\infty )$

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

Now, taking the intersection of solution set for both conditions, we get the solution set as: $m\in (10,\infty )$