Question

Find the value of $m$ such that roots of the quadratic equation $x^2-(m-3)x+m=0$ $(m\in R)$ are negative.

• A. $(-\infty ,1)$
• B. $[1,3)$
• C. $(0,1]$
• D. None of the above

Answer 1

The correct option is C $(0,1]$

Condition : when root are negative,

(i) $D\ge 0$
(ii) Sum of roots $<0$
(iii) Product of roots $>0$


(i) $D\ge 0$

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

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

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

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

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

(ii) Sum of roots $<0\Rightarrow -\frac{b}{a}<0$

$\Rightarrow (m-3)<0\Rightarrow m<3$

$\Rightarrow m\in (-\infty ,3)$

(iii) Product of roots $>0\Rightarrow \frac{c}{a}>0$

$\Rightarrow m>0\Rightarrow m\in (0,\infty )$

Now, taking intersection of all three above condition, we get

$m\in (0,1]$