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# Root | Definition & Meaning

## Definition

When dealing with **function**s, the **root** represents the **value**(s) of the **variable** where a **function** evaluates to zero. For example, the **root** of f(x) = x – 2 is x = 2, since f(x = 2) = 2 – 2 = 0. In some advanced contexts, **roots** are called zeros to avoid confusion with **radicals** (square **roots**, cube **roots**, etc., which are different things entirely).

## What Is a Root?

A **function**‘s **root** is the **variable**‘s **value** that satisfies the equation when the **function** itself equals zero. This concept helps to find the **roots** of the equation, and it becomes easier to solve the equation further. The **roots** of a **function** are also called zeros of a **function**.

**Roots** can be found algebraically as well as graphically. The fundamental theorem of algebra shows that the number of **roots** of a **polynomial** **function** gives the degree of the **polynomial**.

## How To Find the Roots of a Function?

It is a simple method to find **roots** of any type of **function,** whether it is a linear, quadratic, rational, or logarithmic **function**. To find the **roots** of a **function**, we isolate x on one side of the equation and manipulate the equation for the **roots**.

The **function** whose degree is 1 is called a linear **function** or linear **polynomial, **e.g., 5x – 4.

The **function** whose degree is 2 is called a quadratic **function** or quadratic **polynomial,** e.g., x^{2} + 4x + 3.

The **function** of degree 3 is called a cubic **function,** e.g., 2x^{3} + x^{2} – 13x + 6 is a cubic **function**.

Similarly, 16x^{7} + 6x^{4} – 3x^{2} is a **polynomial**/**function** of degree 7.

There is a different method for finding **roots** of different degrees of **function**s.

## Roots of a Linear Function

**Root**/**zero** of a linear **function** is the **value** of the independent **variable** when the dependent **variable** is kept zero, i.e., if an equation has **variables** x and y, then y will be the dependent **variable** and x will be the **independent** **variable,** but if the equation has **variable** other than x and y then the one plotted on vertical axis would be dependent.

To find the zero of the linear **function**, first, we need to identify the dependent and independent **variables**. Then substitute zero for the dependent **variable,** and hence it will drop out of the equation of the **function**. Since the dependent **variable** has now vanished, the equation can be solved for the remaining **variable,** and the solution would be the **root** of the equation of the **function**.

## Roots of a Quadratic Function

Following are the methods to find the **roots** of a quadratic **function**

- Completing square method
- Middle-term break method
- Using the quadratic formula

### Completing Square Method

Consider a quadratic equation:

ax^{2} + bx + c = 0

Rewrite it in the form as:

ax^{2} + bx = -c

Divide the constant **a** into both sides if a is not equal to **1**. Now the equation becomes:

x^{2} + $\dfrac{bx}{a}$ = – $\dfrac{c}{a}$

Divide the x term **coefficient,** i.e., $\left(\dfrac{b}{a}\right)$ by 2 and it becomes $\left(\dfrac{b}{2a}\right)$.

Square it $\left(\dfrac{b}{2a}\right)^2$ and add it to both sides of the equation, now the equation becomes:

x^{2} + $\dfrac{bx}{a}$ + $\left(\dfrac{b}{2a}\right)^2$ Â = – $\dfrac{c}{a}$ + $\left(\dfrac{b}{2a}\right)^2$

The left side of the **equation** can be written as:

$\left(xÂ + \dfrac{b}{2a}\right)^2$

which is the perfect square.

The equation becomes:

$\left(x + \dfrac{b}{2a}\right)^2$ = -$\dfrac{c}{a}$ + $\left(\dfrac{b}{2a}\right)^2$

Now take the square **root** on both sides and solve for the **value** of x. The solution will be the **root** of the equation.

### Middle-term Break Method

Consider a quadratic equation ax^{2} + bx + c = 0 (a = 1). In this method, we split the **coefficient** of x. Think of 2 numbers such that their sum will be equal to b and their product will be equal to c. now factorize the **values** and make a pair of **expressions**.

Equating each expression one by one with zero and extracting out the **values** of x will give the **roots** of the quadratic equation. Since it is an equation of **degree 2** so we will get 2 **roots**.

### Using Quadratic Formula

Consider a quadratic equation ax^{2} + bx + c and take the **values** of coefficients of x^{2}, x, and constant. Put these **values** in the quadratic formula, and the solution will be two **roots** of the quadratic equation.

## Nature of Roots of Quadratic Equation

The nature of **roots** can be specified by finding the **value** of discriminant D.

**Discriminant = D = b ^{2}** –

**4acÂ**

If the **value** of D is greater than zero, i.e., D>0, the **roots** are real and distinct.

If the **value** of D is equal to zero, i.e., D=0, the **roots** are real and equal.

If the **value** of D is less than zero, i.e., D<0, the **roots** are imaginary.

### Roots of a Cubic Function

The **roots** of a cubic **function** can be found by factorizing the given equation into a linear or quadratic equation. Further, the **quadratic or linear equation** can be solved by any of the methods discussed above.

Cubic **polynomials** can be factorized by the **synthetic division** method and remainder theorem.

### Remainder Theorem

In the remainder theorem, guess the **value** for the **variable,** i.e., x. Let us put x = a, y becomes zero; hence (x – a) is a **root** of the cubic **polynomial**. Then we divide the cubic **polynomial** with the **root** (x – a) by long division to obtain a quadratic equation. Further, this quadratic equation can be solved by any of the **three methods** to solve a quadratic equation.

### Synthetic Division Method

In the synthetic division method, guess the **value** of the **variable** which satisfies the equation (same as in the **remainder theorem**). Then write the coefficients of all the **variables** in descending **order** and put **zero** if any of the terms are missing.

Now bring the first term down, multiply it with the **value** of the **root** or divisor, and write it down under the next term. Now add both the **values** and write the answer at the bottom of the row. Repeat the step **3,4** times or as required until you reach the end of the problem. You will get a **quadratic** **equation** that can be easily solved further to find the **roots**.

## Roots of Higher Degree Functions

Roots of higher-degree **polynomials** can also be found by the **remainder** **theorem** and synthetic division method to get a **quadratic** **equation,** and it can be further factorized to find the **roots**.

## Solved Example of Finding Roots

Find the **roots** of the following equations

$\dfrac{x}{2}$ + 3 = 7

x^{2} – 5x + 6 = 0

x^{3} + x^{2} – x – 1 = 0

### Solution

#### Part 1

$\dfrac{x}{2}$ + 3 = 7

$\dfrac{x}{2}$ = 7 – 3

$\dfrac{x}{2}$ = 4

x = 4*2

**x = 8**

Hence 8 is the **root** of the equation.

we can also check if the answer is **correct** or not by putting the answer in the given **equation** as

$\dfrac{8}{2}$ + 3 = 7

4 + 3 = 7

7 = 7

The equation is **satisfied**; hence it is **confirmed** that 8 is the **root**.

#### Part 2

Given x^{2} – 5x + 6 = 0

Using the **middle**–**term** **break** rule,

x^{2} – 2x – 3x + 6 = 0

x(x – 2) – 3(x – 2) = 0

(x – 3)(x – 2) = 0

#### Check

Put x = 3 in the given **equation**.

(3)^{2} – 5(3) + 6 = 0

9 – 15 + 6 = 0

0 = 0

Put x=2 in the given **equation**.

(2)^{2} – 5(2) + 6 = 0

4 – 10 + 6 = 0

0 = 0

Hence, x = 2 and x = 3 are the **roots** of the equation.

#### Part 3

Given x^{3} + x^{2} – x – 1 = 0

Using **synthetic division**, let x = 1 and put it in the **equation**

(1)^{3} + (1)^{2} – 1 â€“ 1 = 0

1 + 1 – 1 – 1 = 0

2 – 2 = 0

0 = 0

x=1 is the **root** of the equation, which can be used in **synthetic division.**

*All images are created using GeoGebra.*