Question

The least positive integral value of $m$ for which the equation $x^2-2(m-1)x+2m+1=0$ has both roots positive is

Answer 1

$x^2-2(m-1)x+2m+1=0$
Since, both roots are positive, then required condions are

$\left(i\right)D\ge 0\Rightarrow (m-1{)}^2-(2m+1)\ge 0$
$\Rightarrow m\in (-\infty ,0]\cup [4,\infty )$

$\left(ii\right)\frac{-b}{a}>0\Rightarrow m>1$

$\left(iii\right)\frac{c}{a}>0\Rightarrow m>-\frac{1}{2}$

$\left(i\right)\cap \left(ii\right)\cap \left(iii\right)\Rightarrow m\ge 4$
So, least positive integer is $4.$