Animated Solution for Mathematics - Quadratic Equations: If both the roots of the quadratic equation $x^2-mx+4=0$ are real and distinct and they lie in the interval $[1,5]$, then $m$ lies in the interval:

Welcome, future engineer. Today, we are not just solving a quadratic equation; we are mastering the art of 'Root Location.' This is a classic JEE Advanced gatekeeper problem.

It separates those who memorize formulas from those who truly visualize the geometry of functions. Imagine you are standing on a vast, flat plain. You have a parabola, $f(x)=x^2-mx+4$, which is a beautiful, upward-opening bowl.

Your mission is to ensure that this bowl dips below the ground (the x-axis) at two distinct points, and crucially, that both of these points occur between the markers $x=1$ and $x=5$.

To solve this, we cannot rely on blind algebraic manipulation. We need a strategy. We need four pillars of logic to hold our roots in place.

Pillar 1: The roots must exist and be distinct. This is the Discriminant condition, $D>0$.

Pillar 2: The parabola must be above the x-axis at the left boundary, $f(1)>0$. (Note: We use strict inequality here because the roots must lie between the markers).

Pillar 3: The parabola must be above the x-axis at the right boundary, $f(5)>0$.

Pillar 4: The vertex (the lowest point of the bowl) must be trapped between $1$ and $5$. If the vertex is outside this range, the roots will be outside, no matter what the boundaries say.

The Discriminant

Let us start with the discriminant, $D=b^2-4ac$. For our equation $x^2-mx+4=0$, we have $a=1$, $b=-m$, and $c=4$.

Plugging these in, we get:

For the roots to be real and distinct, we require $m^2-16>0$. This factors into $(m-4)(m+4)>0$.

Using the wavy curve method, we find that $m$ must be in the interval $(-\infty ,-4)\cup (4,\infty )$. This is our first constraint.

The Boundaries

Now, let us look at the boundaries. For the roots to be trapped between $1$ and $5$, the parabola must be 'above' the x-axis at the edges of our interval.

At $x=1$, we calculate:

Since the parabola must be above the axis, we set $5-m>0$, which gives us $m<5$.

Next, at $x=5$, we calculate:

Setting this $>0$, we get $29>5m$, or $m<5.8$.

The Vertex

Finally, we must ensure the vertex is centered. The x-coordinate of the vertex is given by $-\frac{b}{2a}$.

Substituting our values, we get:

We need this to be strictly between $1$ and $5$, so $1<\frac{m}{2}<5$. Multiplying by $2$, we get $2<m<10$.

Now, we bring it all together. We need the intersection of all our findings: $m\in (-\infty ,-4)\cup (4,\infty )$, $m<5$, $m<5.8$, and $m\in (2,10)$.

Let us look at the number line. The condition $m>4$ combined with $m<5$ gives us the interval $(4,5)$.

The vertex condition $2<m<10$ is already satisfied by this range. Thus, the final solution is $m\in (4,5)$.

You have successfully trapped the roots! This is the power of analytical thinking. You didn't just solve for $x$; you controlled the behavior of the entire function.