Animated Solution for Mathematics - Quadratic Equations: 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 :

Imagine you are standing on a vast, flat plain representing the $x$-axis. You are looking at a parabola defined by the quadratic expression:

The problem asks us to find the integral values of $m$ such that this expression is always positive for any real $x$. Geometrically, this requires the parabola to float entirely above the $x$-axis, never touching or dipping below it.

To ensure our parabola stays forever in the upper half-plane, we must satisfy two mathematical conditions. First, the parabola must open upwards, meaning the leading coefficient must be strictly positive:

Second, the parabola must never touch the $x$-axis, which implies the quadratic equation has no real roots. In algebraic terms, the discriminant $D=b^2-4ac$ must be strictly less than zero:

We begin with the condition $1+2m>0$. Subtracting $1$ from both sides gives $2m>-1$, which simplifies to:

This serves as our primary boundary condition for the variable $m$.

Now, we evaluate the discriminant $D<0$. Substituting the coefficients into the formula, we have:

Dividing the entire inequality by $4$ simplifies the expression to:

Expanding the binomials, we obtain:

Combining like terms leads to the elegant quadratic inequality:

To solve $m^2-6m-3<0$, we find the roots of $m^2-6m-3=0$ using the quadratic formula:

Given that $\sqrt{3}\approx 1.732$, the roots are approximately $3-3.464=-0.464$ and $3+3.464=6.464$. The inequality holds for $m$ in the interval:

Intersecting this with our first condition, $m>-0.5$, we find that the valid range remains $(-0.464,6.464)$. The integers contained within this interval are $0,1,2,3,4,5,$ and $6$.

There are exactly 7 such integral values of $m$.