A **polynomial** whose highest exponent is 2 is called a quadratic expression.

The standard form of a quadratic expression is:

**ax ^{2} + bx + c**

(in this polynomial, the highest power of the variable x is 2)

The standard form of a quadratic equation is:

ax

In the quadratic expression (or the equation),

**Example 1:**

**3√x ^{2} – 5x + 4 = 0** is not a quadratic equation as √x

**Example 2:**

**x(x + 3) = x(x – 3) + 9** is also not a quadratic equation.

On simplifying, the equation becomes:

**x ^{2} + 3x = x^{2} – 3x + 9, i.e. 6x –9 = 0**, again the highest exponent in x is not 2

**Example 3:**

**x ^{2} + ^{1}/_{x}^{2} = 1** is also not a quadratic equation.

The power of the variable x changes to 4 on simplifying:

Roots are values of x for which the quadratic equation becomes equal to 0.

**Example:**

**x ^{2} – 5x + 6 = 0** is a quadratic equation that becomes 0 on writing 2 or 3 in

i.e. 22 – 5 × 2 + 6 = 0, 4 – 10 + 6 = 0, 10 – 10 = 0 or

32 – 5 × 3 + 6 = 0, 9 – 15 + 6 = 0, 15 – 15 = 0.

Therefore, 2 or 3 are roots of the quadratic equation

**2. Number of roots of a quadratic equation:**

A quadratic equation can have at most 2 roots, i.e. one or two roots.

No quadratic equation can have more than 2 roots.

**Tip:** Substitute values of x in the equation to check if the quadratic reduces to 0. If yes, the particular value is root of the quadratic.

**Example 1:**

Is 2 a root of the quadratic: **x ^{2} + 5x – 14 = 0**

Solution: plug 2 in x in the equation. We get:

So, 2 is a root of the quadratic.

**Example 2:**

Is k a root of the quadratic: **x ^{2} + kx – 2k^{2}?**

Substitute k in x:

so, k is a root of the given quadratic.

**3. Solving a Quadratic Equation:**

To solve a quadratic equation is to find roots of the quadratic, i.e. to find zeroes of the quadratic.

There are four methods to solve a quadratic equation:

1. Factorization method2. Substitute and Factorize

3. By completing the square

4. Quadratic formula method

Let us discuss the four methods of finding roots of a quadratic:

**Example 1:**

Solve the quadratic: **x ^{2} – 5x + 6 = 0**

x

x (x – 2) – 3(x – 2) = 0

(x – 3)(x – 2) = 0.

Now, the product

Either

x – 3 = 0 or x – 2 = 0, i.e.

x = 3 or x = 2.

So, we say the roots of the given quadratic are

**V. Imp Note:**

Use the connective “or” for the two roots of a quadratic but not “and”,

for the product **(x – 3) (x – 2)** to be 0 any of the two **(x – 3), (x – 2)** factors can be 0, whereas

“and” lays the condition that both roots must be 0 at the same time for the product to be 0, which is not necessary.

**Example 2:**

**Solve the quadratic:** **6x ^{2} – 5x + 1 = 0**

Now find two numbers whose:

Product is:

So, 6x

6x

2x (3x – 1) – 1(3x – 1) = 0,

(2x – 1)(3x – 1) = 0,

So, either 2x – 1 = 0 or 3x – 1= 0, i.e.

2x = 1, x = ½ or 3x = 1, x = 1/3

The roots are

Let √x = y, then x = y2. Now, the quadratic is:

4y – y2 = 3,

4y – y2 – 3 = 0,

Reverse the signs of all terms to make leading coefficient positive and to factorize:

y

y

y(y – 3) – 1(y – 3) = 0,

(y – 1)(y – 3) = 0, so

y – 1= 0 or y – 3 = 0, so

y = 1 or y = 3.

Now, replace y = √x, so

√x = 1, x = 1 or

√x = 3, x = 9

Solution: Substitute

Use the two laws of exponents:

to simplify the given quadratic as below:

3

Now let y = 3

y × 3

9y +

9y

9y

9y

9y(y – 1) – 1(y – 1) = 0,

(9y – 1)(y – 1) = 0,

9y – 1= 0 or y – 1= 0,

y =

Now replace y by 3x , so

3

3

3

**To solve a quadratic with completing the square method, you will add and deduct: ^{1}/_{4}(coefficient of x)^{2}**

**E****xample 1: Solve the quadratic: x ^{2} + 8x + 4 = 0**

Solution: You cannot find two numbers whose product is 4 and sum is 8.

So, we will use completing the square method to solve the given quadratic.

Express the given quadratic in the form of the identity:

x

(x

so, add

**Important Tip:**

**To solve a quadratic with completing the square method,
You will add and deduct: ^{1}/_{4} (coefficient of x)^{2}**

Subtract

Now, the quadratic:

x

(x + 4)

(x + 4)

(x + 4)

x + 4 = ± √ (12), so,

x + 4 = √ 12, or x + 4 = - √ 12, so

x = 4 + √ 12 or x = 4 – √ 12

**Example 2:**

**Solve the quadratic: 2x ^{2} + x – 4= 0**

(2) × (- 4) i.e. – 8 and whose sum is +1. Use the completing square method.

1. Divide the quadratic with the leading coefficient (2, here)

2. Add and subtract

Now add and subtract

The quadratic becomes: x

Write

x

x

(x +

(x +

x +

x =

x =

Consider the standard quadratic equation:

The roots of a quadratic equation written in standard form are:

x =

x =

(We will discuss the proof later on)

1. x

Solution:

a = 1, b = –5, c = 6.

√ (b

Now, the two roots are:

x =

x =

**2. 3x ^{2} + 11x – 4= 0**

b

√ (b

Now, the roots are:

x =

x =