**What you will learn in this lesson on Quadratic Equations?**

1. What is a quadratic equation?

A polynomial in which the highest exponent is 2 is called a quadratic equation
The standard form of a quadratic expression is ax^{2} + bx + c

Equate the quadratic expression to 0 and you get the standard form of the quadratic equation:

ax^{2} + bx + c = 0

in the quadratic equation, a is the leading coefficient.

a and b are called coefficients of x^{2} and x, while c is called constant term.

What are not quadratic expressions?

Example 1:

√x^{2} – 5x + 6 is not a quadratic equation, as the highest exponent of x is not 2 for √x^{2} is same as x, in which the exponent is 1 but not 2

**Example 2:**

(x – 2) (x – 3) = (x + 1) (x – 4) is also not a quadratic equation as the highest exponent of x on simplifying is not 2, but 1.

**Example 3:**

x^{2} + ^{1}/ _{x}^{2} = 0 is not a quadratic equation, as the highest exponent of the polynomial is 4 and not 2.

**Example 4:**

x^{2} + 2√x + 1 = 0 is not a quadratic equation, as the power of x, in the second term 2√x, is ^{1}/_{2}, not an integer. (See definition of a Polynomial)

2. Roots of a quadratic equation

Roots of a quadratic equation are values of the variable for which the *quadratic expression* becomes equal to 0.

**Example 1:**

Consider the quadratic equation: x^{2} – 2x + 1 = 0.

If x = 1, then the quadratic expression x^{2} – 2x + 1 becomes: 1^{2} – 2(1) + 1 = 0.

This value 1 of the variable x, for which the quadratic equation reduces to 0 is called root of the quadratic equation: x^{2} – 2x + 1 = 0.

Example 2:

Consider the quadratic equation: x^{2} – 5x + 6 = 0

Plug 2 or 3 in x and the quadratic expression x^{2} – 5x + 6 becomes equal to 0.

So, 2 or 3 are roots of the quadratic equation x^{2} – 5x + 6 = 0

*Note: Any quadratic equation can have at most two roots, i.e. one or two roots, but not more than *2.

1. In example 1 above, x = 1 is the only root.

When both the roots of a quadratic equation are equal, we call the root a “double root”.
In example 1 above, 1 is a double root

2. in example 2 above, x = 2 or x = 3 are the two roots.

In this quadratic equation, the two roots are real and different.

3. How to solve a quadratic equation:

To solve a quadratic equation is to find its roots.

We will discuss four methods of solving a quadratic equation, i.e. finding roots. They are:

1. Factorization method

2. Substitute and Factorize

3. By completing the square

4. Quadratic formula method

Let us capture a brief overview of each method:

Example 1: Solve the quadratic x^{2} – 3x + 2 = 0

Solution: You must be thorough with factoring methods learnt in factoring lesson.

Factorize: x^{2} – 2x – x + 2 = 0,

x(x – 2) – 1(x – 2) = 0,

(x – 2) (x – 1) = 0,

So, x – 2 = 0 or x – 1 = 0,

x = 2 or x = 1 are the two roots of the given quadratic equation.

in short, 2 or 1 are the roots

Note: Use “or” to connect the two roots.

Do not use “and” to connect the two roots.

Do not say 2 “and” 1 are roots.

Why? See content for quadratic equations

**2. Substitute and Factorize:
Example 1: **

** Solve the quadratic equation: 5 ^{1 + x} + 5 ^{1 – x} = 26**

Solution: first simplify the quadratic as:

5 × 5^{x} + ^{5}/_{5}^{x} = 26

Substitute y for 5x.

Now the quadratic is: 5y + ^{5}/_{y} = 26,

5y^{2} + 5 = 26y, i.e.

5y^{2} – 26y + 5 = 0,

5y^{2} – 25y –y + 5 = 0,

5y(y – 5) – 1(y – 5) = 0,

(5y – 1) (y – 5) = 0,

5y – 1 = 0 or y – 5 = 0,

y = 1/5 or y = 5,

Now put 5^{x} = ^{1}/_{5} or 5^{x} = 5,

5^{x} = 5 ^{– 1} or 5^{x} = 5,

x = - 1 or x = 1.

**3. Completing the square method:
Example 1: x**

Solution: add 25 to each side to complete the square on the left side:

x^{2} + 10x + 25 = 75 + 25

x^{2} + 10x + 25 = 100,

(x + 5)^{2} = 102,

x + 5 = 10 or x + 5 = -10

x = 5 or x = -15

**4. Using the quadratic formula: **

For the quadratic equation ax^{2} + bx + c = 0, the two roots are:

x = ^{[–b – √ (b2 – 4ac)]}/_{2a},

x = ^{[–b + √ (b2 – 4ac)]}/_{2a}

**Example 1: **

** Find the roots of the quadratic equation x ^{2} – 9x + 36 = 0 using quadratic formula. **

Solution:

In the quadratic x^{2} – 13x + 36 = 0:

a = 1, b = - (-13) = 13 and c = 36,

b^{2} – 4ac = 132 – 4 × 1 × 36 = 169 – 144 = 25

so, of the two roots, one root is :

x = ^{[–b–√ (b2 – 4ac)]}/_{2a},

x = ^{[13 – √ (25)]}/_{2} ×1 = ^{[13 – 5 ]}/_{2} = ^{8}/_{2} = 4

and the other root is:

x = ^{[–b + √ (b2 – 4ac)]}/_{2a}

x = ^{[13 + √ (25)]}/_{2} ×1 = ^{[13 + 5]}/_{2} = ^{18}/_{2} = 9

**5. Nature of roots of a quadratic equation using the discriminant**

By nature of roots is meant whether roots are real or complex and equal or different.

b^{2} – 4ac is called discriminant of a quadratic equation.

We use discriminant to find the nature of roots of a quadratic equation.

**6. Sum and Product of roots of a quadratic equation:**

In the quadratic equation ax^{2} + bx + c = 0

1. Sum of the roots = - (^{b}/_{2}) = - ^{(coefficient of x)}/ _{(coefficient of x2)}

Example1:

In the quadratic equation: 3x^{2} – 12x + 36 = 0,

Sum of the roots is – ^{(-12)}/_{3} = 4

2. Product of roots = ^{c}/_{a} = ^{(constant term)} / _{(leading coefficient)}

[Leading coefficient is same as coefficient of x^{2}]

Example 2:

In the quadratic equation: 3x^{2} – 12x + 36 = 0,

Product of the roots is ^{36}/_{3} = 12

**Example 1: In the quadratic equation 9x ^{2} – 12x + 36 = 0, **

Sum of the roots is – (- ^{12}/_{9}) = ^{12}/_{9} = ^{4}/_{3} and

Product of roots is ^{36}/_{12} = 3.

**7. Signs of roots of a quadratic equation: **

Consider the quadratic equation ax^{2} + bx + c = 0

**How to solve simultaneous linear equation**