Question

$x^2-(m-3)x+m=0(m\in R)$ be a quadratic equation. Find the value of $m$ for which, at least one root is greater than $2$.

• A. $[-9,9]$
• B. $[9,\infty )$
• C. $(-\infty ,9]$
• D. None of the above.

Answer 1

The correct option is B $[9,\infty )$

$f\left(x\right)=x^2-(m-3)x+m=0,mϵR$

$△={(m-3)}^2-4m\ge 0$ ($\because$ at least one root)

$m^2-10m+9\ge 0$

$\therefore mϵ(-\infty ,1]\bigcup [9,\infty )$

For $m<1$, both roots are always $<2$

For example, $m=0$

$x^2+3x,x=0,-3$ (Both are $<2$)

$\therefore$For $mϵ[9,\infty ]$ at least one root is greater than $2$.