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 smaller than $2$

• A. $(-\frac{3}{2},3)$
• B. $(-\infty ,1]$
• C. $(0,4]$
• D. $(1,\infty )$

Answer 1

The correct option is B $(-\infty ,1]$

$f\left(x\right)=x^2-(m-3)x+m=0bothrootsarelessthan2\therefore f\left(2\right)>0f\left(2\right)=2^2-(m-3)2+m>04-2m+6+m>010-m>0m-10<0m\in (-\infty ,10)\&D\ge 0{(m-3)}^2-4m\ge 0m^2-6m+9-4m\ge 0m^2-10m+9\ge 0(m-1)(m-9)\ge 0m\in (-\infty ,1)U(9,\infty )\frac{m-3}{2}<2\left(conditionforvertex\right)m<4+3m<7m\in (-\infty ,7)\therefore m\in (-\infty ,1)$