Animated Solution for Mathematics - Quadratic Equations: The smallest value of $k$, for which both the roots of the equation $x^2-8kx+16(k^2-k+1)=0$ are real, distinct and have values at least 4, is .........

A Journey into the Parabola

Imagine you are standing on a vast, flat plane, and before you, a parabola $f(x)=x^2-8kx+16(k^2-k+1)$ begins to emerge. This isn't just an algebraic expression; it is a dynamic shape that shifts and stretches as you change the parameter $k$.

Our mission today is to pin this parabola down. We are told that this curve must intersect the x-axis at two distinct points, and both of those points must lie at or to the right of the value $x=4$. This is a classic 'Location of Roots' problem—a beautiful intersection of algebra and geometry.

To force this parabola to behave exactly as we want, we cannot rely on luck. We need a rigorous framework. Think of these as the three pillars that hold up our solution:

1. The Reality Check ($D>0$): We need the roots to be real and distinct. If the discriminant $D$ were negative, the parabola would float above the x-axis, never touching it. If $D=0$, the roots would be identical. Thus, we demand $D=b^2-4ac>0$.

2. The Symmetry Constraint ($-\frac{b}{2a}>4$): The vertex of our parabola is the axis of symmetry. If this axis is to the left of 4, the roots will likely fall on the wrong side of our boundary. We must ensure the vertex is firmly positioned to the right of $x=4$.

3. The Boundary Guard ($f(4)\ge 0$): This is the most subtle condition. Even if the vertex is to the right of 4, if the parabola dips below the x-axis before it reaches 4, one of our roots will be 'lost' to the left side. By ensuring $f(4)\ge 0$, we guarantee the parabola is still above the x-axis at the boundary.

Let us tackle the first pillar. The discriminant is given by:

Simplifying this, we get $64k^2-64(k^2-k+1)>0$. Dividing by 64, the $k^2$ terms vanish, leaving us with $k-1>0$, or simply $k>1$. This is our first gatekeeper.

Next, the vertex position:

Setting $4k>4$ gives us $k>1$. It is fascinating how the geometry aligns here; the first two conditions yield the same result.

Finally, the boundary guard. We evaluate $f(4)=(4{)}^2-8k(4)+16(k^2-k+1)$. Expanding this, we find:

This simplifies to $16k^2-48k+32\ge 0$. Dividing by 16, we arrive at $k^2-3k+2\ge 0$. Factoring this quadratic, we get $(k-1)(k-2)\ge 0$. Using the wavy curve method, we see that $k$ must be in the interval $(-\infty ,1]\cup [2,\infty )$.

Now, we bring it all together. We need $k>1$ AND $k\in (-\infty ,1]\cup [2,\infty )$.

When we intersect these sets, the region $(-\infty ,1]$ is discarded because it contradicts $k>1$. We are left with the elegant result: $k\ge 2$.

The smallest value of $k$ that satisfies all these conditions is, therefore, 2. You have successfully navigated the constraints, balanced the geometry, and arrived at the truth. Remember, in JEE Advanced, it is rarely about brute force; it is about understanding the 'soul' of the equation.