Animated Solution for Mathematics - Quadratic Equations: The real numbers $x_1,x_2,x_3$ satisfying the equation $x^3-x^2+\beta x+\gamma =0$ are in AP. Find the intervals in which $\beta$ and $\gamma$ lie.

Imagine you are standing before a cubic equation, $x^3-x^2+\beta x+\gamma =0$. At first glance, it looks like a standard polynomial, but there is a hidden rhythm here—a secret structure.

We are told that the roots $x_1,x_2,x_3$ are in an Arithmetic Progression (A.P.). In the world of competitive mathematics, whenever you see an A.P., you should immediately think of symmetry.

Instead of labeling our roots as $x,x+d,x+2d$, which would lead to a messy algebraic tangle, let us be more clever. Let us define our roots as $x_1=a-d$, $x_2=a$, and $x_3=a+d$.

Why do this? Because when we invoke Vieta's formulas, the sum of the roots becomes a beautiful, simple expression: $(a-d)+a+(a+d)=3a$.

According to Vieta, the sum of the roots is equal to the negative of the coefficient of $x^2$ divided by the coefficient of $x^3$. In our equation, that is $-(-1)/1=1$.

Thus, we have the elegant result $3a=1$, or $a=1/3$. We have just pinned down the middle root without breaking a sweat!

Since $a=1/3$ is a root, it must satisfy the original equation. Substituting this into $x^3-x^2+\beta x+\gamma =0$, we get:

Simplifying this, we find $\frac{1}{27}-\frac{1}{9}+\frac{\beta }{3}+\gamma =0$, which reduces to $-\frac{2}{27}+\frac{\beta }{3}+\gamma =0$.

This gives us a vital bridge: $\gamma =\frac{2}{27}-\frac{\beta }{3}$. This equation tells us that $\beta$ and $\gamma$ are not independent; they are locked in a dance. If you know one, you know the other.

Now, we turn to the second Vieta relation: the sum of the products of the roots taken two at a time, which equals $\beta$. Using our symmetric roots, we calculate:

Expanding this, we get $(a^2-ad)+(a^2+ad)+(a^2-d^2)=\beta$, which simplifies to $3a^2-d^2=\beta$. Substituting our known value $a=1/3$, we arrive at $3(1/9)-d^2=\beta$, or $1/3-d^2=\beta$.

Here is the moment of truth: for the roots to be real, the common difference $d$ must be a real number. Therefore, $d^2$ must be greater than or equal to zero.

This implies $1/3-\beta \ge 0$, or $\beta \le 1/3$.

With $\beta \le 1/3$ in our pocket, we return to our bridge equation $\gamma =\frac{2}{27}-\frac{\beta }{3}$. As $\beta$ decreases, $\gamma$ must increase.

By substituting the maximum value of $\beta =1/3$ into this relation, we find the minimum value for $\gamma$:

And there we have it! The intervals are $\beta \in (-\infty ,1/3]$ and $\gamma \in [-1/27,\infty )$.

You have successfully navigated the cubic landscape by using symmetry to simplify the complex. This is the heart of JEE Advanced mathematics—not brute force, but the elegant application of fundamental truths.