Question

$x^2-(m-3)x+m=0(m\in R)$ be a quadratic equation. Find the value of $m$ for which, both the roots lie in the interval of $(1,2)$

• A. $(10,\infty )$
• B. $(-\infty ,0)$
• C. $(-\infty ,\infty )$
• D. None of the above.

Answer 1

The correct option is D None of the above.

$f\left(x\right)=x^2-(m-3)x+m=0rootsliebetween(1,2)1.D\ge 0{(m-3)}^2-4m\ge 0m^2-6m+9-4m\ge 0m^2-10m+9\ge 0(m-9)(m-1)\ge 0m\in (-\infty ,1)U(9,\infty )2.f\left(1\right)>01-m+3+m>04>0\therefore m\in R3.f\left(2\right)>02^2-2m+6+m>010-m>0m-10<0m\in (-\infty ,10)4.1<\frac{m-3}{2}<22<m-3<45<m<7m\in (5,7)\therefore novalueofmexists$