The number of integral values 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

We are given the quadratic expression

$$f(x)= (1+2m)x^{2}-2(1+3m)x+4(1+m), \qquad x\in \mathbb R.$$

For every real value of $$x$$ the value of $$f(x)$$ must be positive. For a quadratic $$ax^{2}+bx+c$$ the following well-known facts hold:

• If $$a>0$$ and the discriminant $$D=b^{2}-4ac<0,$$ the parabola opens upward and has no real roots, so the quadratic is strictly positive for all $$x\in\mathbb R$$. • If $$a\le 0,$$ or if $$a>0$$ but $$D\ge 0,$$ the quadratic will touch or cross the $$x$$-axis somewhere, so it cannot stay positive everywhere.

Hence we must impose simultaneously

$$a=1+2m>0 \quad\text{and}\quad D<0.$$

Now we calculate the discriminant in detail. Here

$$a=1+2m,\qquad b=-2(1+3m),\qquad c=4(1+m).$$

First compute $$b^{2}$$:

$$b^{2}=\left[-2(1+3m)\right]^{2}=4(1+3m)^{2}=4\bigl(1+6m+9m^{2}\bigr).$$

Next compute the product $$4ac$$:

$$4ac=4\bigl(1+2m\bigr)\cdot 4(1+m)=16(1+2m)(1+m).$$

So the discriminant is

$$D=b^{2}-4ac=4\bigl(1+6m+9m^{2}\bigr)-16(1+2m)(1+m).$$

We factor the common $$4$$ out:

$$D=4\Bigl[(1+6m+9m^{2})-4(1+2m)(1+m)\Bigr].$$

Let us expand the bracket inside the big parentheses. First expand $$(1+2m)(1+m):$$

$$(1+2m)(1+m)=1(1+m)+2m(1+m)=1+m+2m+2m^{2}=1+3m+2m^{2}.$$

Multiply by $$4$$:

$$4(1+2m)(1+m)=4\bigl(1+3m+2m^{2}\bigr)=4+12m+8m^{2}.$$

Now perform the subtraction:

$$1+6m+9m^{2}-\bigl(4+12m+8m^{2}\bigr) =1-4+6m-12m+9m^{2}-8m^{2} =(-3)-6m+m^{2} =m^{2}-6m-3.$$

Therefore

$$D=4\bigl(m^{2}-6m-3\bigr).$$

For the quadratic to be always positive we need $$D<0$$, so

$$4\bigl(m^{2}-6m-3\bigr)<0 \;\Longrightarrow\; m^{2}-6m-3<0.$$

This is a quadratic inequality. First find its roots using the quadratic formula:

$$m=\frac{6\pm\sqrt{(-6)^{2}-4\cdot1\cdot(-3)}}{2} =\frac{6\pm\sqrt{36+12}}{2} =\frac{6\pm\sqrt{48}}{2} =\frac{6\pm4\sqrt3}{2} =3\pm2\sqrt3.$$

The coefficient of $$m^{2}$$ is positive, so the parabola opens upward and the expression $$m^{2}-6m-3$$ is negative between its roots. Hence

$$3-2\sqrt3<m<3+2\sqrt3.$$

We must also satisfy the earlier condition $$1+2m>0$$, i.e.

$$m>-\,\frac12.$$

Notice that

$$3-2\sqrt3\approx3-3.464\approx-0.464,$$

and $$-\,\dfrac12=-0.5$$ is slightly smaller. Therefore the tighter lower bound is $$m>3-2\sqrt3\;(\approx-0.464).$$ Combining the two conditions gives the single open interval

$$3-2\sqrt3<m<3+2\sqrt3\;\;(\approx-0.464<m<6.464).$$

The integral (integer) values of $$m$$ lying strictly inside this interval are

$$0,\,1,\,2,\,3,\,4,\,5,\,6.$$

That is a total of $$7$$ integers.

Hence, the correct answer is Option A.