Question

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

• A. $(9,10)$
• B. $(10,\infty )$
• C. None of the above.
• D. $(1,10)$

Answer 1

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

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


Condition :

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

now, solving it,

(i) $D>0$

$\Rightarrow (m-3{)}^2-4m>0$

$\Rightarrow m^2-6m+9-4m>0$

$\Rightarrow m^2-10m+9>0$

$\Rightarrow (m-1)(m-9)>0$

$m\in (-\infty ,1)\cup (9,\infty )$

(ii) $f\left(2\right)<0$

$\Rightarrow (2{)}^2-(m-3\left)\right(2)+m<0$

$\Rightarrow 4-2m+6+m<0$

$\Rightarrow -m+10<0\Rightarrow m>10$

$m\in (10,\infty )$

Now, taking intersection of both condition, we get

$m\in (10,\infty )$