Question

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

• A. $(-\infty ,1)$
• B. $(-\infty ,1)\cup (9,\infty )$
• C. $(-\infty ,0)$
• D. $(-\infty ,9)$

Answer 1

The correct option is C $(-\infty ,0)$

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 $a=1,b=-(m-3),c=m.$

Applicable conditions are
$\left(i\right)D>0$
$\left(ii\right)$ Product of root $<0$

Taking $\left(i\right)D>0$
$\Rightarrow (-(m-3){)}^2-4\cdot 1\cdot m>0$
$\Rightarrow m^2-6m+9-4m>0$
$\Rightarrow m^2-10m+9>0$
$\Rightarrow (m-1)(m-9)>0$
$m\in (-\infty ,1)\cup (9,\infty )$

$\left(ii\right)$ Product of roots $<0$
$\Rightarrow \frac{m}{1}<0\Rightarrow m<0$
$m\in (-\infty ,0)$

Now, taking intersection of both the solution sets, we get:
$m\in \left\{\right(-\infty ,1)\cup (9,\infty \left)\right\}\cap (-\infty ,0)$
$m\in (-\infty ,0)$