Question

The Quadratic equation ${x^2} + px + q = 0$ has a complex root $2 + 3i$ . Find p and q.

Answer 1

Hint: Quadratic equations is a polynomial equation of a second degree, which implies that it comprises a minimum of one term that is squared. The power of variable x are always non-negative integers, hence the equation is a polynomial equation with highest power as 2. The solution for this equation is the values of x, which are also called zeroes.
To solve this question we will use the sum of roots and the product of roots. As we know complex numbers always are in pairs. So, we can find another root and solve it.

Complete step by step answer:
Suppose the roots of the general quadratic equation:
$a{x^2} + bx + c = 0$ are $\alpha $ and $\beta $ , then using the root properties we have:
Sum of roots = $\alpha + \beta $ = $\dfrac{{ - b}}{a}$
product of roots= $\alpha \beta $ = $\dfrac{c}{a}$
Complex roots always appear in conjugate pairs, so if one root of the given quadratic is $\alpha = 2 + 3i$ then the other root is $\beta = 2 - 3i$
${x^2} + px + q$
We know that,
$\alpha + \beta = \dfrac{{ - p}}{1}$ ……(i)
$\alpha \beta = \dfrac{q}{1}$ …….(ii)
We can calculate
$\alpha + \beta = (2 + 3i) + (2 - 3i)$
$\alpha + \beta = 4$
From equation (i) $p = - 4$
$\alpha \beta = (2 + 3i)(2 - 3i)$
Applying the identity of $(a + b)(a - b) = {a^2} - {b^2}$
$\alpha \beta = {(2)^2} - {(3i)^2}$
As we know ${i^2} = - 1$
$\alpha \beta = 4 + 9$
$\alpha \beta = 13$
From equation (ii) $q = 13$
Therefore the value p and q are $ - 4$ and $13$

Note:
A quadratic equation with real or complex coefficients has two solutions, called roots. These two solutions may or may not be distinct, and they may or may not be real. In the quadratic formula, the expression underneath the square root sign is called the discriminant of the quadratic equation, and is often represented using D
$D = {b^2} - 4ac$
So the value of D defines the nature of roots whether they are real or not.