Animated Solution for Mathematics - Quadratic Equations: If both the roots of the quadratic equation $x^2-2kx+k^2+k-5=0$ are less than 5, then $k$ lies in the interval

A Journey into Quadratic Elegance

Welcome, future engineer! Today, we are not just solving an equation; we are sculpting a parabola. When you look at the quadratic equation $x^2-2kx+k^2+k-5=0$, I want you to see more than just symbols.

Because the coefficient of $x^2$ is $1$, which is positive, we know immediately that this parabola opens upwards. It is a beautiful, symmetric U-shape.

The problem asks us to ensure that both roots of this equation are less than $5$. This is a classic 'Location of Roots' problem, a favorite in JEE Advanced because it tests your ability to translate algebraic constraints into geometric reality.

To force this parabola to have both its roots to the left of the vertical line $x=5$, we cannot rely on just one condition. We need three distinct pillars of logic.

First, the roots must exist in the real plane. If the discriminant $D$ were negative, the parabola would float entirely above the x-axis, and we would have no roots to talk about. So, our first pillar is $D\ge 0$.

Second, consider the vertex. The vertex is the horizontal center of our parabola. If the vertex were to the right of $5$, it would be impossible for both roots to be to the left of $5$. Thus, our second pillar is $-\frac{b}{2a}<5$.

Finally, consider the boundary. Even if the vertex is to the left of $5$, the parabola could still cross the x-axis after $5$. To prevent this, we must ensure that at $x=5$, the parabola is still 'flying high' above the x-axis. Thus, our third pillar is $f(5)>0$.

Let us build these pillars one by one. For the first pillar, the discriminant $D=b^2-4ac$. Substituting $a=1$, $b=-2k$, and $c=k^2+k-5$, we get:

Expanding this, we find $4k^2-4k^2-4k+20\ge 0$. Notice how the $4k^2$ terms cancel out? We are left with $-4k+20\ge 0$, which simplifies beautifully to $k\le 5$.

Now, for the second pillar, the vertex position. The x-coordinate of the vertex is $-\frac{b}{2a}$. Substituting our values, we get:

This is a strict inequality, reinforcing our first condition.

Finally, the third pillar: the boundary value $f(5)>0$. We substitute $x=5$ into our function $f(x)=x^2-2kx+k^2+k-5$. This gives us:

Combining like terms, we arrive at $k^2-9k+20>0$. Factorizing this quadratic, we get $(k-4)(k-5)>0$. Using the wavy curve method, we find that $k$ must be in the interval $(-\infty ,4)\cup (5,\infty )$.

We have our three conditions: $k\le 5$, $k<5$, and $k\in (-\infty ,4)\cup (5,\infty )$. To find the final answer, we must find the intersection of these sets.

The regions $k<5$ and $k\le 5$ are satisfied by $k<5$. When we intersect this with $(-\infty ,4)\cup (5,\infty )$, the only part that survives is $k<4$.

And there you have it! The interval is $(-\infty ,4)$. You have successfully navigated the geometry of the parabola and constrained it to your will. Keep this mindset—visualize the geometry, build your pillars, and the algebra will always follow.