Animated Solution for Mathematics - Quadratic Equations: Let ${\alpha }_1,{\alpha }_2,\dots ,{\alpha }_7$ be the roots of the equation $x^7+3x^5-13x^3-15x=0$ and $|{\alpha }_1|\ge |{\alpha }_2|\ge \cdots \ge |{\alpha }_7|$. Then ${\alpha }_1{\alpha }_2-{\alpha }_3{\alpha }_4+{\alpha }_5{\alpha }_6$ is equal to ______.

Welcome, future engineer. Today, we are not just solving a polynomial; we are unmasking a hidden symmetry.

When you first look at the equation $x^7+3x^5-13x^3-15x=0$, it is natural to feel intimidated by the degree 7. But in the world of JEE Advanced, a high-degree polynomial is often a mask hiding a much simpler face.

Factoring the Oddity

Look closely at the equation. Every single term contains an $x$. This is our first clue.

By factoring out $x$, we get:

Immediately, we have found our first root: $x=0$. We have reduced the problem from a degree 7 to a degree 6.

But look at the remaining polynomial: $x^6+3x^4-13x^2-15=0$. Notice that every power of $x$ is even. This is not a coincidence; it is an invitation.

Whenever you see only even powers, think of it as a 'biquadratic' structure. Let us introduce a new variable, $t=x^2$.

Suddenly, our terrifying degree 6 equation transforms into a friendly cubic:

This is the heart of the problem. We need to find the roots of this cubic. Using the rational root theorem or simple trial and error, we test values.

If we plug in $t=3$, we get $27+27-39-15=0$. It works! So, $(t-3)$ is a factor.

Dividing the cubic by $(t-3)$ leaves us with $(t^2+6t+5)=0$, which factors beautifully into $(t+5)(t+1)=0$. Our values for $t$ are $3,-1,-5$.

Now, we must return to $x$. Since $x^2=t$, we have three scenarios:

1. For $t=3$, $x=\pm \sqrt{3}$. 2. For $t=-1$, $x=\pm i$. 3. For $t=-5$, $x=\pm i\sqrt{5}$.

Including our initial root $x=0$, we have our seven roots: $0,\pm \sqrt{3},\pm i,\pm i\sqrt{5}$.

To order them by magnitude, we visualize them on the complex plane. The magnitude $|z|$ is simply the distance from the origin.

We have $|0|=0$, $|\pm i|=1$, $|\pm \sqrt{3}|=\sqrt{3}\approx 1.732$, and $|\pm i\sqrt{5}|=\sqrt{5}\approx 2.236$.

Sorting these, we assign our $\alpha$ values: ${\alpha }_1,{\alpha }_2=\pm i\sqrt{5}$, ${\alpha }_3,{\alpha }_4=\pm \sqrt{3}$, ${\alpha }_5,{\alpha }_6=\pm i$, and ${\alpha }_7=0$.

Now, we evaluate the expression ${\alpha }_1{\alpha }_2-{\alpha }_3{\alpha }_4+{\alpha }_5{\alpha }_6$.

First, ${\alpha }_1{\alpha }_2=(i\sqrt{5})(-i\sqrt{5})=-i^2(5)=5$.

Second, ${\alpha }_3{\alpha }_4=(\sqrt{3})(-\sqrt{3})=-3$.

Third, ${\alpha }_5{\alpha }_6=(i)(-i)=-i^2=1$.

Putting it all together:

There it is. A beautiful, clean integer result emerging from the chaos of complex roots.

Remember, in JEE Advanced, the math is rarely about brute force; it is about finding the right substitution to reveal the underlying order. Keep practicing, and keep looking for that hidden symmetry.