Animated Solution for Mathematics - Quadratic Equations: If $2+i\sqrt{3}$ is a root of the equation $x^2+px+q=0$, where $p$ and $q$ are real, then $(p,q)=(\dots ,\dots )$.

Welcome, future engineers! Today, we are going to unravel a classic JEE problem that tests your understanding of the deep, elegant connection between complex numbers and polynomial equations.

We are given the quadratic equation $x^2+px+q=0$ and told that one of its roots is $\alpha =2+i\sqrt{3}$. We are also given the crucial information that $p$ and $q$ are real numbers.

This is the master key to the entire problem. Imagine the complex plane where our root $\alpha$ sits at the coordinate $(2,\sqrt{3})$. Because the coefficients $p$ and $q$ are real, the polynomial is constrained by a beautiful symmetry known as the Complex Conjugate Root Theorem.

In the world of polynomials with real coefficients, complex roots never travel alone. They always appear in conjugate pairs, acting like a mirror reflection across the real axis.

If $a+ib$ is a root, then its reflection, $a-ib$, must also be a root. This occurs because evaluating the polynomial at the conjugate causes the imaginary parts to cancel out perfectly, leaving a result of zero.

So, if our first root is $\alpha =2+i\sqrt{3}$, our second root, $\beta$, must be its conjugate:

The Bridge

Now that we have identified both roots, we need to find the coefficients $p$ and $q$. Vieta's formulas serve as the bridge between the roots of a polynomial and its coefficients.

For a quadratic equation of the form $x^2+px+q=0$, the relationships are defined as:

First, let us find the sum of the roots:

Since the sum of the roots is $-p$, we have $-p=4$, which implies $p=-4$.

Next, we calculate the product of the roots using the difference of squares identity:

Recalling that $i^2=-1$, the expression simplifies as follows:

Since the product of the roots is $q$, we find $q=7$. We have arrived at our destination: $(p,q)=(-4,7)$.