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Learn when quadratic equations have imaginary roots, how to identify them using discriminant, and solve complex numbers one step at a time with examples.

---

## What are Non-Real Roots?

Quadratic equations $$ax^2 + bx + c = 0$$ sometimes have solutions that cannot be found in ordinary numbers. These solutions are called "non-real roots" or "imaginary roots."

Visible text: Quadratic equations sometimes have solutions that cannot be found in ordinary numbers. These solutions are called "non-real roots" or "imaginary roots."

Imagine we're looking for a number that, when multiplied by itself, gives a negative result. Does such a number exist? No! Because any number multiplied by itself always gives a positive result or zero. This is where the concept of imaginary numbers begins.

## Imaginary Numbers

Imaginary numbers are numbers that contain $$i$$, where $$i = \sqrt{-1}$$. This means $$i^2 = -1$$.

Visible text: Imaginary numbers are numbers that contain , where . This means .

Examples of imaginary numbers:

- $$3i$$ (read as: "three i")
- $$2 + 5i$$ (read as: "two plus five i")
- $$-4i$$ (read as: "negative four i")

Visible text: - (read as: "three i")
- (read as: "two plus five i")
- (read as: "negative four i")

Numbers like $$2 + 5i$$ are called complex numbers, because they are a combination of a real number $$2$$ and an imaginary number $$5i$$.

Visible text: Numbers like are called complex numbers, because they are a combination of a real number and an imaginary number .

## When Does a Quadratic Equation Have Imaginary Roots?

A quadratic equation has imaginary roots when its discriminant is negative. The discriminant is $$D = b^2 - 4ac$$.

Visible text: A quadratic equation has imaginary roots when its discriminant is negative. The discriminant is .

If $$D < 0$$, then the quadratic equation will have two different imaginary roots.

Visible text: If , then the quadratic equation will have two different imaginary roots.

## How to Find Imaginary Roots

To find imaginary roots, we still use the formula:

Component: MathContainer
Children:

```math
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
```

```math
x = \frac{-b \pm \sqrt{-(4ac - b^2)}}{2a}
```

```math
x = \frac{-b \pm i\sqrt{4ac - b^2}}{2a}
```

### First Example Problem

Let's find the roots of the equation $$x^2 + 4x + 5 = 0$$.

Visible text: Let's find the roots of the equation .

**Step** $$1$$: Identify the values of $$a$$, $$b$$, and $$c$$.

Visible text: **Step** : Identify the values of , , and .

- $$a = 1$$
- $$b = 4$$
- $$c = 5$$

Visible text: - 
- 
-

**Step** $$2$$: Calculate the discriminant.

Visible text: **Step** : Calculate the discriminant.

```math
D = b^2 - 4ac = 4^2 - 4 \cdot 1 \cdot 5 = 16 - 20 = -4
```

Since $$D = -4 < 0$$, this equation has imaginary roots.

Visible text: Since , this equation has imaginary roots.

**Step** $$3$$: Use the quadratic formula.

Visible text: **Step** : Use the quadratic formula.

Component: MathContainer
Children:

```math
x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-4 \pm \sqrt{-4}}{2 \cdot 1}
```

```math
x = \frac{-4 \pm 2i}{2} = -2 \pm i
```

Therefore, the roots of the equation $$x^2 + 4x + 5 = 0$$ are $$x_1 = -2 + i$$ and $$x_2 = -2 - i$$.

Visible text: Therefore, the roots of the equation are and .

### Second Example Problem

Determine the type of roots for the equation $$2x^2 + x + 3 = 0$$.

Visible text: Determine the type of roots for the equation .

**Step** $$1$$: Identify the values of $$a$$, $$b$$, and $$c$$.

Visible text: **Step** : Identify the values of , , and .

- $$a = 2$$
- $$b = 1$$
- $$c = 3$$

Visible text: - 
- 
-

**Step** $$2$$: Calculate the discriminant.

Visible text: **Step** : Calculate the discriminant.

```math
D = b^2 - 4ac = 1^2 - 4 \cdot 2 \cdot 3 = 1 - 24 = -23
```

Since $$D = -23 < 0$$, this equation has imaginary roots.

Visible text: Since , this equation has imaginary roots.

**Step** $$3$$: Find the equation's roots.

Visible text: **Step** : Find the equation's roots.

Component: MathContainer
Children:

```math
x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-1 \pm \sqrt{-23}}{2 \cdot 2}
```

```math
x = \frac{-1 \pm i\sqrt{23}}{4}
```

Therefore, the roots of the equation are $$x_1 = \frac{-1 + i\sqrt{23}}{4}$$ and $$x_2 = \frac{-1 - i\sqrt{23}}{4}$$.

Visible text: Therefore, the roots of the equation are and .

## Why Do Imaginary Roots Always Come in Pairs?

Imaginary roots always appear in pairs in the form of $$a + bi$$ and $$a - bi$$. These pairs are called "complex conjugates."

Visible text: Imaginary roots always appear in pairs in the form of and . These pairs are called "complex conjugates."

This happens because the quadratic formula involves $$\pm\sqrt{D}$$. When $$D < 0$$, we get $$\pm i\sqrt{|D|}$$, which gives us complex conjugate pairs.

Visible text: This happens because the quadratic formula involves . When , we get , which gives us complex conjugate pairs.