Question

$x^2-(m-3)x+m=0(m\in R)$ be a quadratic equation. Find the value of $m$ for which both roots are real and distinct:

• A. $(-1,1)$
• B. $(-\infty ,-1)U(9,\infty )$
• C. $(-\infty ,-4)U(12,\infty )$
• D. $(-4,3)U(5,\infty )$

Answer 1

The correct option is B $(-\infty ,-1)U(9,\infty )$

Given quadratic equation is $x^2-(m-3)x+m=0$.

Also given that both roots are real and distinct.

We have to find the value of $m$.

Since the roots are real and distinct, the discriminant $b^2-4ac$ is greater than $0$.

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

We have $b^2-4ac>0$

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

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

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

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

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

Computing signs of each term we get

$m-9=0\Rightarrow m=9,m-9<0\Rightarrow m<9,m-9>0\Rightarrow m>9$

$m-1=0\Rightarrow m=1,m-1<0\Rightarrow m<1,m-1>0\Rightarrow m>1$

Summarize the signs in a table we get

	$m<1$	$m=1$	$1<m<9$	$m=9$	$m>9$
$m-9$	$-$	$-$	$-$	$0$	$+$
$m-1$	$-$	$0$	$+$	$+$	$+$
$(m-9)(m-1)$	$+$	$0$	$-$	$0$	$+$

Thus $(m-9)(m-1)>0$ when $m<1$ and $m>9$ .

Therefore the value of $m$ is $(-\infty ,1)\cup (9,\infty )$.