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

Answer 1

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

If roots are opposite in sign , then we can say $0$ lies between the roots of $x^2-(m-3)x+m=0$.
Applicable conditions are
(i) $D>0$ and
(ii) Product of root $<0$

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

Now, taking intersection of both the condition, we get
$m\in (-\infty ,0)$