| Content On This Page | ||
|---|---|---|
| Solving Quadratic Equations with Real Coefficients yielding Complex Roots | ||
Quadratic Equations with Complex Roots
Solving Quadratic Equations with Real Coefficients yielding Complex Roots
When the Discriminant is Negative
A standard quadratic equation in one variable is expressed in the form $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are coefficients representing real numbers, and $a$ is non-zero ($a \neq 0$). The nature of the roots (or solutions) of this equation is entirely determined by a value called the discriminant, which is calculated as $\Delta = b^2 - 4ac$.
Based on the value of the discriminant, we can predict the type of roots the quadratic equation will have:
- If $\Delta > 0$: The equation has two distinct real roots. These roots can be rational or irrational.
- If $\Delta = 0$: The equation has exactly one real root. This root is often referred to as a repeated root or a root with multiplicity 2.
- If $\Delta < 0$: The equation has no real roots. Instead, it has two distinct roots which are complex numbers.
This section specifically addresses the scenario where the coefficients $a, b,$ and $c$ are real numbers, but the discriminant $\Delta = b^2 - 4ac$ turns out to be negative. In such cases, the quadratic equation $ax^2 + bx + c = 0$ has complex roots.
Using the Quadratic Formula to Find Complex Roots
The most reliable and general method to find the roots of a quadratic equation $ax^2 + bx + c = 0$, regardless of the nature of the roots, is by using the quadratic formula. The formula is:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $
(The Quadratic Formula)
When the discriminant $\Delta = b^2 - 4ac$ is negative, the term under the square root, $\sqrt{b^2 - 4ac}$, involves the square root of a negative number. Let $D = b^2 - 4ac$. Since $D < 0$, we can write $D$ as $D = -k$, where $k$ is a positive real number. Specifically, $k = -(b^2 - 4ac) = 4ac - b^2$, which is positive.
Now, let's look at $\sqrt{D}$ when $D$ is negative:
$$ \sqrt{D} = \sqrt{-k} $$Using the property of square roots and the definition of the imaginary unit $i = \sqrt{-1}$ ($i^2 = -1$), where $k > 0$:
$$ \sqrt{-k} = \sqrt{k \times (-1)} = \sqrt{k} \times \sqrt{-1} $$ $$ \sqrt{-k} = \sqrt{k} \cdot i $$In our case, $k = -(b^2 - 4ac) = 4ac - b^2$. So, $\sqrt{b^2 - 4ac} = \sqrt{-(4ac - b^2)} = i \sqrt{4ac - b^2}$. Note that $\sqrt{4ac - b^2}$ is a positive real number since $4ac - b^2 > 0$.
Substituting this expression for $\sqrt{b^2 - 4ac}$ into the quadratic formula when $\Delta < 0$:
$$ x = \frac{-b \pm i \sqrt{4ac - b^2}}{2a} $$This formula clearly shows that the roots are complex numbers because they involve the imaginary unit $i$. The '$\pm$' sign indicates that there are two distinct roots:
The first root is $x_1 = \frac{-b + i \sqrt{4ac - b^2}}{2a}$.
We can write this as a complex number in the standard form $P + Qi$:
$$ x_1 = \frac{-b}{2a} + \frac{\sqrt{4ac - b^2}}{2a} i $$The second root is $x_2 = \frac{-b - i \sqrt{4ac - b^2}}{2a}$.
Writing this in the standard form $P - Qi$:
$$ x_2 = \frac{-b}{2a} - \frac{\sqrt{4ac - b^2}}{2a} i $$Let $P = \frac{-b}{2a}$ and $Q = \frac{\sqrt{4ac - b^2}}{2a}$. Since $a, b, c$ are real numbers and $4ac - b^2 > 0$, $P$ and $Q$ are real numbers. Also, since $4ac - b^2 > 0$, $Q \neq 0$.
The two roots are therefore $x_1 = P + Qi$ and $x_2 = P - Qi$. This shows that when a quadratic equation with real coefficients has complex roots, they always appear as a pair of complex conjugate roots. The complex conjugate of a complex number $a+bi$ is $a-bi$.
Examples of Solving Quadratic Equations with Complex Roots
Example 1. Solve the equation $x^2 + x + 1 = 0$.
Answer:
The given equation is $x^2 + x + 1 = 0$.
This is a quadratic equation in the standard form $ax^2 + bx + c = 0$.
Comparing the coefficients, we have $a=1$, $b=1$, and $c=1$. These are all real coefficients.
First, calculate the discriminant $D = b^2 - 4ac$:
$$ D = (1)^2 - 4(1)(1) = 1 - 4 = -3 $$Since $D = -3$ is negative ($D < 0$), the roots of the equation are complex.
Now, apply the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$:
$$ x = \frac{-(1) \pm \sqrt{-3}}{2(1)} $$ $$ x = \frac{-1 \pm \sqrt{(-1) \times 3}}{2} $$Using $\sqrt{-1} = i$:
$$ x = \frac{-1 \pm i\sqrt{3}}{2} $$The two distinct roots are obtained by taking the '+' and '-' signs:
$$ x_1 = \frac{-1 + i\sqrt{3}}{2} $$ $$ x_2 = \frac{-1 - i\sqrt{3}}{2} $$We can write these roots in the standard form $a+bi$:
$$ x_1 = -\frac{1}{2} + \frac{\sqrt{3}}{2} i $$ $$ x_2 = -\frac{1}{2} - \frac{\sqrt{3}}{2} i $$These roots are indeed a pair of complex conjugates.
Answer: The roots are $-\frac{1}{2} + \frac{\sqrt{3}}{2} i$ and $-\frac{1}{2} - \frac{\sqrt{3}}{2} i$.
Example 2. Solve the equation $3x^2 - 2x + 1 = 0$.
Answer:
The given equation is $3x^2 - 2x + 1 = 0$.
Comparing with $ax^2 + bx + c = 0$, we have $a=3$, $b=-2$, and $c=1$. These are all real coefficients.
Calculate the discriminant $D = b^2 - 4ac$:
$$ D = (-2)^2 - 4(3)(1) = 4 - 12 = -8 $$Since $D = -8$ is negative ($D < 0$), the roots of the equation are complex.
Now, apply the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$:
$$ x = \frac{-(-2) \pm \sqrt{-8}}{2(3)} $$ $$ x = \frac{2 \pm \sqrt{(-1) \times 8}}{6} $$Using $\sqrt{-1} = i$ and simplifying $\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}$:
$$ x = \frac{2 \pm i \sqrt{8}}{6} = \frac{2 \pm i (2\sqrt{2})}{6} $$ $$ x = \frac{2 \pm 2\sqrt{2}i}{6} $$Now, we can simplify the fraction by dividing the numerator and denominator by the common factor $2$:
$$ x = \frac{\cancel{2}(1 \pm \sqrt{2}i)}{\cancel{6}^{3}} $$ $$ x = \frac{1 \pm \sqrt{2}i}{3} $$The two distinct roots are:
$$ x_1 = \frac{1 + \sqrt{2}i}{3} $$ $$ x_2 = \frac{1 - \sqrt{2}i}{3} $$Writing these roots in the standard form $a+bi$:
$$ x_1 = \frac{1}{3} + \frac{\sqrt{2}}{3} i $$ $$ x_2 = \frac{1}{3} - \frac{\sqrt{2}}{3} i $$These roots are a pair of complex conjugates.
Answer: The roots are $\frac{1}{3} + \frac{\sqrt{2}}{3} i$ and $\frac{1}{3} - \frac{\sqrt{2}}{3} i$.
Property: Complex Roots Occur in Conjugate Pairs
A very significant property in the study of polynomials with real coefficients is the Conjugate Root Theorem. It states that if a polynomial equation with **real coefficients** has a complex root of the form $a+bi$ where $b \neq 0$ (i.e., a non-real complex root), then its complex conjugate $a-bi$ must also be a root of that polynomial.
As demonstrated by the quadratic formula when the discriminant is negative ($D < 0$), the roots are always found in the form $\frac{-b}{2a} \pm \frac{\sqrt{4ac - b^2}}{2a}i$. This structure directly produces the two roots as a complex conjugate pair, confirming this theorem for quadratic equations with real coefficients.
This property has several implications, including the fact that any polynomial equation with real coefficients and an odd degree must have at least one real root, because complex roots must occur in pairs.
Understanding how to solve quadratic equations with negative discriminants introduces us to complex numbers as solutions to algebraic problems, expanding our ability to find roots for a wider range of equations.