Question

If $x^2-(a-3)x+a=0$ has atleast one positive root, then $a\in$

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

Answer 1

The correct option is A $(-\infty ,0)\cup [9,\infty )$

Real roots exist when $D\ge 0$
$\Rightarrow (a-3{)}^2-4a\ge 0$
$\Rightarrow a^2-10a+a\ge 0$
$\Rightarrow (a-9)(a-1)\ge 0$
$\Rightarrow a\in (-\infty ,1]\cup [9,\infty )$

Case $1:$ Both roots are positive
$\Rightarrow D\ge 0,a-3>0,a>0$
$\Rightarrow a\in (-\infty ,1]\cup [9,\infty ),a>3,a>0$
$\Rightarrow a\in [9,\infty )\cdots \left(1\right)$

Case $2:$ When exactly one root is positive
$\Rightarrow D\ge 0,a<0$
$\Rightarrow a\in (-\infty ,1]\cup [9,\infty ),a<0$
$\Rightarrow a\in (-\infty ,0)\cdots \left(2\right)$

When one root is $0$, then $a=0$
$x^2+3x=0\Rightarrow x=0,-3$
So, $a=0$ not possible.

From $\left(1\right)$ and $\left(2\right)$
$\Rightarrow a\in (-\infty ,0)\cup [9,\infty )$