Question

The number of integral value(s) of $m$ for which the quadratic expression,
$(1+2m)x^2-2(1+3m)x+4(1+m),x\in R,$ is always positive, is

Answer 1

The given equation is always positive if
$1+2m>0\text{and}\Delta <0$
$1+2m>0\Rightarrow m>-\frac{1}{2}\dots \left(1\right)$

Now, $\Delta <0$
$\Rightarrow 4(1+3m{)}^2-16(1+2m\left)\right(1+m)<0$
$\Rightarrow 1+9m^2+6m-4-12m-8m^2<0$
$\Rightarrow m^2-6m-3<0\Rightarrow (m-3{)}^2<12\Rightarrow -2\sqrt{3}<m-3<2\sqrt{3}$
$\Rightarrow 3-2\sqrt{3}<m<3+2\sqrt{3}$
$\Rightarrow m\in (3-\sqrt{12},3+\sqrt{12})\dots \left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$, integral value of
$m=0,1,2,3,4,5,6$
Hence, number of integral values of $m$ is $7$.