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Solution of Quadratic Equations in the Complex Number System

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How to Solve Quadratic Equations with Complex Roots: Key Techniques

A complex number is such a numeral that consists of 2 parts: one that is real and another that is imaginary. It can be written by way of a + bi where a and b stands for real numbers (including 0) and i is an imaginary number.

                          a       +        bi

                         ↑                 ↑

                         Real Part        Imaginary Part

 

The Prowess of “i” in Complex Equations

Under the concept of the real number system, there is Absolutely 'no' solution to the equation \[x^{2}\] = -1x.


However, you can still have a solution to the equation using a new number. Learning about this new number system will allow you to know that the equation does have a solution. The cornerstone of this new number system is the numeral iii, also called as the imaginary unit.


i²= -1i = −1i, squared, equals, minus, \[1\sqrt{-1}\] = −1i ​= i square root of, (-), 1, end square root, (=), i


This suggests that by taking multiples of this imaginary unit, we can bring into being innumerable many more new numbers, like i, \[i\sqrt{5}i\], 3i 3i, -12i − 12i minus, 12, i.


These are some of the examples of imaginary numbers.


That being said, you can even go ahead by adding real numbers and imaginary numbers, for example, 2+7i²+7i², plus, 7, i, and 3 -\[\sqrt{2}\] i³ - and combination of these is called complex numbers.

 

How Complex Numbers Help to Solve Quadratic Equations by Using the Quadratic Formula?

Solving Quadratics with complex numbers as the Solution is not as complex as it sounds to be. When a problem demands you for the roots, it is just a similar thing as asking for the zeros or the x-intercepts. These are the points where y = 0, so we can replace that value to start with.


Taking into account, we're performing this example to know all about complex equations or complex number solutions. It's all going to decide the discriminant. For example, given a quadratic expression in its usual form (y = ax² + bx + c), the discriminant is b² - 4ac. In case the discriminant is positive, you'll obtain two real answers. If it comes to be equivalent to zero, you're only collecting one real answer. However, if the discriminant is negative, that's when you obtain two complex solutions to the problem.


This is because of the way the quadratic formula functions. If you attempt to solve the problem incorporating the quadratic formula, you might observe that the discriminant is a component of the formula that is in the interior of the square root.


Here, if you substitute a, b and c into the formula and then start to assess the equation, you will notice the discriminant will, unquestionably, become negative. You will finish with \[-\sqrt{31}\], which will be an imaginary number. Hence, it’s logical to state if that part is negative, you are going to get an imaginary number, closing the answer to a complex number.

 

Quadratic Equations & Formula

A quadratic equation is an equation that can be written in a form of ax² + bx + c = 0, or an equation that we are able to minimize or rearrange to this standard form. The equation brings out two solutions which may be alike or unlike. Thus, the most optimal manner to solve a quadratic equation is to use the quadratic formula.


In this equation, x is indicative of an unknown number. In addition, a, b, and c are indicative of the known numbers.


Quadratic Equation and Number of Solutions it Has

A quadratic equation is a mathematical expression of degree 2. The answer to How many solutions does any quadratic equation can have is given by the Fundamental Theorem of Algebra:


To take note of, “A polynomial equation has no less than 1 solution”


Any equation that is written in a standard form of a1x (n-1) + a2x (n-2) + a3x (n-3) + … + an = 0 is a polynomial equation of degree ‘n’. When applies the fundamental theorem of algebra to polynomial equations of this form, renders a significant solution, which is:

“A polynomial equation of degree ‘n’, owns ‘n’ roots”. Thus, a quadratic equation that is a polynomial equation of order 2 has two solutions. Let us see further.


A common form of the quadratic equation can be such as ax² + bx + c = 0. Solving it for x, we obtain the below two solutions:


x = \[\frac{-b\pm \sqrt{b^{2} - 4ac}}{2a}\]


\[2a - b\pm (b^{2} - 4ac) \frac{1}{2}\]


This is called the quadratic formula and provides two values for ‘x’. One for the (plus{+}) sign and the other for the (minus {–}) sign. The volume in the square root is known as the discriminant D. We have the following three terms to it:


In the case of D = 0, the ± sign not anymore plays a role and we get a single, real solution, repeated twice.


If D > 0, the volume in the square root is positive and hence the solution becomes real. In this case, there are two real solutions that are different.


However, if D < 0, the volume in the square root is negative and thus complex or imaginary numbers.


Now that you have a thorough knowledge of complex numbers and quadratic expressions, you can use them to solve any quadratic equations with D < 0.


Fun facts 

  1. Complex numbers are actually an addition to the real number system

  2. The addition of complex numbers makes a significant difference in mathematics

  3. Complex numbers help in the solution of quadratic equations with the help of the Quadratic formula

  4. When solving any quadratic equations, each complex result will always have his conjugate companion with him— which are known as complex conjugates.

  5. Despite the fact that complex numbers have an imaginary part, there are actually a number of real-life applications of these "imaginary" numbers, such as wavering springs and voltaic electronics.

  6. Imaginary numbers are used by the engineers to explain electric current and its intensity in the real world.

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FAQs on Solution of Quadratic Equations in the Complex Number System

1. How do you solve a quadratic equation that results in complex roots?

To solve a quadratic equation of the form ax² + bx + c = 0 that has complex roots, you follow these steps using the quadratic formula:

  • Step 1: Identify the coefficients 'a', 'b', and 'c' from your equation.
  • Step 2: Calculate the discriminant (D = b² - 4ac). If the discriminant is negative (D < 0), the equation will have complex roots.
  • Step 3: Substitute 'a', 'b', and 'c' into the quadratic formula: x = [-b ± √(b² - 4ac)] / 2a.
  • Step 4: Simplify the square root of the negative discriminant by expressing it in terms of the imaginary unit 'i', where √(-k) = i√k for any positive number k.
  • Step 5: The two complex roots will be a pair of complex conjugates.

2. What role does the discriminant (b² - 4ac) play in finding complex solutions to a quadratic equation?

The discriminant, D = b² - 4ac, is a critical component of the quadratic formula that determines the nature of the roots of an equation with real coefficients without having to fully solve it. Specifically for complex solutions:

  • If D > 0, there are two distinct real roots.
  • If D = 0, there is exactly one real root (a repeated root).
  • If D < 0, there are no real roots. Instead, the equation has two complex conjugate roots. A negative discriminant is the definitive indicator that the solutions will involve the imaginary unit 'i'.

3. Why do complex roots of a quadratic equation with real coefficients always appear in conjugate pairs?

This is a fundamental property based on the structure of the quadratic formula, x = [-b ± √(b² - 4ac)] / 2a. When the coefficients a, b, and c are real numbers and the discriminant is negative, the term √(b² - 4ac) produces a purely imaginary number, say 'iy'. The formula then becomes x = (-b ± iy) / 2a. This expression naturally splits into two roots: one with a '+iy' term and one with a '-iy' term. These two roots, (-b/2a) + i(y/2a) and (-b/2a) - i(y/2a), are by definition complex conjugates of each other.

4. When do we need to use the complex number system to solve a quadratic equation?

We must use the complex number system to find solutions for a quadratic equation whenever its discriminant (b² - 4ac) is less than zero. In the real number system, taking the square root of a negative number is undefined. However, the Fundamental Theorem of Algebra states that a polynomial of degree 'n' must have 'n' roots. To satisfy this theorem for quadratic equations with a negative discriminant, we extend our number system to include complex numbers, using the imaginary unit i = √-1 to express the roots completely.

5. What is the maximum number of complex solutions a quadratic equation can have?

According to the Fundamental Theorem of Algebra, a quadratic equation (which is a polynomial of degree 2) will always have exactly two roots in the complex number system. These two roots can be:

  • Two distinct real numbers (e.g., x=2, x=3).
  • One repeated real number (e.g., x=4, x=4).
  • A pair of non-real complex conjugates (e.g., x = 2 + 3i, x = 2 - 3i).
In all cases, the total count of roots is two.

6. What is the geometric interpretation when a quadratic equation has complex roots?

Geometrically, the real roots of a quadratic equation y = ax² + bx + c are the x-coordinates where its graph (a parabola) intersects the x-axis. When a quadratic equation has complex roots, its corresponding parabola does not intersect the x-axis at all. If the coefficient 'a' is positive, the parabola's vertex is entirely above the x-axis. If 'a' is negative, the parabola's vertex is entirely below the x-axis. The complex roots exist in the complex plane (Argand plane) but have no corresponding points on the real number line (x-axis).

7. What is an example of solving a quadratic equation with complex roots?

Let's solve the equation x² + 4x + 5 = 0.

  • 1. Identify coefficients: a = 1, b = 4, c = 5.
  • 2. Find the discriminant: D = b² - 4ac = 4² - 4(1)(5) = 16 - 20 = -4. Since D is negative, the roots are complex.
  • 3. Apply the quadratic formula: x = [-4 ± √(-4)] / 2(1).
  • 4. Simplify the imaginary part: x = [-4 ± 2i] / 2.
  • 5. Find the conjugate roots: x = -2 ± i. The two solutions are x₁ = -2 + i and x₂ = -2 - i.