Question

If $x^2-(a-3)x+a=0$ has atleast one positive root then 'a' belongs to

• A. $\left(-\infty ,0\right)\cup \left[7,9\right]$
• B. $\left(-\infty ,0\right)\cup \left[7,\infty \right)$
• C. $\left(-\infty ,1\right]\cup \left[9,\infty \right)$
• D. $\left(-\infty ,\infty \right)$

Answer 1

Answer: (C)

$\left(-\infty ,1\right]\cup \left[9,\infty \right)$

$\begin{array}{c}{x}^2-\left(a-3\right)x+a=0\\ D\ge 0\\ \Rightarrow {\left(a-3\right)}^2-4\left(1\right)\left(a\right)\ge 0\\ \Rightarrow a^2-6a+9-4a\ge 0\\ \Rightarrow a^2-10a+9\ge 0\\ \Rightarrow \left(a-9\right)\left(a-1\right)\ge 0\\ a\le 1anda\ge 9\end{array}$