Animated Solution for Mathematics - Quadratic Equations: Let $a,b,c$ be real. If $a{x}^2+bx+c=0$ has two real roots $\alpha$ and $\beta$, where $\alpha <-1$ and $\beta >1$, then show that $1+c/a+|b/a|<0$.

Welcome, future engineers. Today, we are going to peel back the layers of a classic JEE Advanced problem. It is not just about solving for roots; it is about understanding the soul of a quadratic equation.

We are given $a{x}^2+bx+c=0$ with roots $\alpha$ and $\beta$, where $\alpha <-1$ and $\beta >1$. Our goal is to prove that $1+\frac{c}{a}+\left|\frac{b}{a}\right|<0$. Let us begin.

Normalization and the Smile

First, let us look at our equation: $a{x}^2+bx+c=0$. It feels a bit cluttered, so let us simplify our life by dividing the entire equation by $a$.

We define a new function:

By doing this, we have normalized the leading coefficient to $1$. Because the leading coefficient is positive, our parabola opens upwards—like a smile. This is the first step in our visualization; we are now dealing with a defined, upward-opening parabola.

The Trap of the Roots

Now, let us place our roots. We are told $\alpha <-1$ and $\beta >1$. Imagine the $x$-axis; we have $\alpha$ on the far left and $\beta$ on the far right.

The points $-1$ and $1$ are trapped strictly between them. Geometrically, this means that the interval $(-1,1)$ is a subset of the interval $(\alpha ,\beta )$.

Recall the property of an upward-opening parabola: between its roots, the function dips below the $x$-axis. This means for any $x$ such that $\alpha <x<\beta$, the function value $f(x)$ must be negative.

Since $-1$ and $1$ are trapped in this region, we immediately know that $f(-1)<0$ and $f(1)<0$. This is the geometric key that unlocks the entire algebraic proof.

The Algebraic Bridge

Let us translate this geometry into algebra. We substitute $x=1$ into our function $f(x)=x^2+\frac{b}{a}x+\frac{c}{a}$:

Since $f(1)<0$, we have our first inequality: $1+\frac{c}{a}+\frac{b}{a}<0$.

Next, we substitute $x=-1$:

Since $f(-1)<0$, we have our second inequality: $1+\frac{c}{a}-\frac{b}{a}<0$.

The Elegant Conclusion

We now have two powerful statements: $1+\frac{c}{a}+\frac{b}{a}<0$ and $1+\frac{c}{a}-\frac{b}{a}<0$. Look closely at these; both expressions share the term $1+\frac{c}{a}$.

In the first, we add $\frac{b}{a}$; in the second, we subtract it. If both results are negative, then the maximum of these two results must also be negative.

The maximum of $\frac{b}{a}$ and $-\frac{b}{a}$ is defined as the absolute value, $\left|\frac{b}{a}\right|$. Therefore, it follows logically that:

We have arrived at our destination. It is a beautiful synthesis of geometry and algebra, and that, my friends, is the essence of JEE Advanced mathematics.