91Ó°ÊÓ

A quadratic equation is a polynomial equation of degree 2. It has the general form:
\[ ax^2 + bx + c =0 \]
where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable. The solutions to a quadratic equation are found using the quadratic formula:
\[ x = \frac{-b \, \pm \, \sqrt{b^2 - 4ac}}{2a} \]
Here:

• \(a\) is the coefficient of \(x^2\)
• \(b\) is the coefficient of \(x\)
• \(c\) is the constant term.

With this formula, we can find the roots of any quadratic equation by calculating the values of \(x\).

The discriminant of a quadratic equation provides essential information about the nature of the roots. It is denoted by \( \Delta \) (delta) and is calculated using the formula:
\[ \Delta = b^2 - 4ac \]
This value can tell us whether the roots are real or complex:

• If \( \Delta > 0 \), the quadratic equation has two distinct real roots.
• If \( \Delta = 0 \), the quadratic equation has exactly one real root (or a repeated real root).
• If \( \Delta < 0 \), the quadratic equation has two complex roots (conjugates).

In our exercise, the discriminant \( \Delta = -12 \), which is less than zero, indicating that the roots are complex.

When the discriminant of a quadratic equation is negative, the roots are complex numbers. A complex number has the form:
\[ a + bi \]
where:

• \( a \) is the real part
• \( bi \) is the imaginary part with \( i \), the imaginary unit, defined as \( i^2 = -1 \).

Using the quadratic formula for our equation \( t^2 - 2t + 4 = 0 \), we get:
\[ t = \frac{2 \, \pm \sqrt{-12}}{2} \]
Simplifying this, we find the complex roots:
\[ t = 1 \, \pm \sqrt{3}i \]
These roots are in the form of complex conjugates: \( 1 + \sqrt{3}i \) and \( 1 - \sqrt{3}i \). Complex roots always appear in conjugate pairs when solving polynomial equations with real coefficients.

In solving higher-degree polynomials, like the given equation \( z^6 - 2z^3 + 4 = 0 \), we can use substitution to simplify it.
By letting \( u = z^3 \), the polynomial reduces to a quadratic form:
\[ u^2 - 2u + 4 = 0 \]
Using the roots from the earlier quadratic equation \( t^2 - 2t + 4 = 0 \), we find \( u = 1 + \sqrt{3}i \) and \( u = 1 - \sqrt{3}i \).
Substituting back for \( z \), we solve:

• \( z^3 = 1 + \sqrt{3}i \)
• \( z^3 = 1 - \sqrt{3}i \)

Each of these cubic equations has three roots, giving us a total of six solutions for \( z \). They can be expressed using polar coordinates and the De Moivre's Theorem:
For \( z^3 = 1 + \sqrt{3}i \), we find:
\[ 2^{1/3} (\cos(\frac{\pi}{9} + \frac{2k\pi}{3}) + i \sin(\frac{\pi}{9} + \frac{2k\pi}{3})) \]
where \( k = 0, 1, 2 \). Similarly, we find the roots for \( z^3 = 1 - \sqrt{3}i \) using:
\[ 2^{1/3} ( \cos( -\frac{\pi}{9} + \frac{2k\pi}{3}) + i \sin( -\frac{\pi}{9} + \frac{2k\pi}{3}) ) . \]
These steps help to break down the polynomial into simpler parts to find all possible roots.