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

The standard form of a quadratic expression is:

ax2 + bx + c
(in this polynomial, the highest power of the variable x is 2)
The standard form of a quadratic equation is:
ax2 + bx + c = 0
In the quadratic expression (or the equation), a, b and c are called constants.

## Examples of what are not quadratic equations:

Example 1:
3√x2 – 5x + 4 = 0 is not a quadratic equation as √x2 becomes x on simplifying and the highest exponent is not 2, then.

Example 2:
x(x + 3) = x(x – 3) + 9 is also not a quadratic equation.
On simplifying, the equation becomes:
x2 + 3x = x2 – 3x + 9, i.e. 6x –9 = 0, again the highest exponent in x is not 2

Example 3:
x2 + 1/x2 = 1 is also not a quadratic equation.
The power of the variable x changes to 4 on simplifying:
x4 –2x + 1 = 0

## 1. Roots of a Quadratic Equation:

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

Example:
x2 – 5x + 6 = 0 is a quadratic equation that becomes 0 on writing 2 or 3 in x.
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 x2 – 5x + 6 = 0.
Roots of a quadratic equation are also called zeroes of the quadratic, because the value of the graph is 0 at each root.

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: x2 + 5x – 14 = 0
Solution: plug 2 in x in the equation. We get:
22 + 5(2) – 14 = 4 + 10 – 14 = 0.
So, 2 is a root of the quadratic.

Example 2:
Is k a root of the quadratic: x2 + kx – 2k2?
Solution:
Substitute k in x:
k2 + k(k) – 2k2 = 2k2 – 2k2 = 0
so, k is a root of the given quadratic.

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

4. Methods of Solving a Quadratic Equation:
There are four methods to solve a quadratic equation:
1. Factorization method
2. Substitute and Factorize
3. By completing the square

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

## 1. Factorization Method:

Example 1:
Solve the quadratic: x2 – 5x + 6 = 0
Solution: factorize: x2 – 5x + 6 = 0 as follows:
x2 – 2x – 3x + 6 = 0
x (x – 2) – 3(x – 2) = 0
(x – 3)(x – 2) = 0.

Now, the product (x – 3) (x – 2) is equal to 0, if
Either (x – 3) is 0 or (x – 2) is 0, i.e.
x – 3 = 0 or x – 2 = 0, i.e.
x = 3 or x = 2.

So, we say the roots of the given quadratic are 2 or 3.

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: 6x2 – 5x + 1 = 0

Solution: (compare the given quadratic with ax2 + bx + c).
Now find two numbers whose:
Product is: a × c, i.e. 6 × 1 and Sum is the middle term i.e. “b” which is –1
So, 6x2 – 5x + 1 = 0,
6x2 – 2x – 3x + 1= 0,
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 are1/2 or 1/3

## 2. Substitute and Factorize Method:

1. Solve the quadratic: 4√x – x = 3

Solution: Substitute y for vx, i.e.
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:
y2 – 4y + 3 = 0,
y2 – 3y – y + 3 = 0,
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

2. 3x + 2 + 3 –x = 10
Solution: Substitute 3x = y
Use the two laws of exponents:
am + n = am × an and a –m = 1/am
to simplify the given quadratic as below:
3x × 32 + 1/3x = 10
Now let y = 3x, so
y × 32 + 1/y = 10, so
9y + 1/y = 10, so
9y2 + 1 = 10y, so
9y2 – 10y + 1 = 0,
9y2 – 9y – y + 1 = 0,
9y(y – 1) – 1(y – 1) = 0,
(9y – 1)(y – 1) = 0,
9y – 1= 0 or y – 1= 0,
y = 1/9, y = 1,

Now replace y by 3x , so
3x = 1/9 or 3x = 1,
3x = 3 –2, so x = - 2, or
3x = 1, so x = 0 [since 30 = 1]

## 3. Completing the Square Method:

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

Example 1: Solve the quadratic: x2 + 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: a2 + 2ab + b2
x2 + 8x + 4 = 0,
(x2 + 2 × x × 4 + 42) – 42 + 4 = 0

so, add 42 to x2 + 8x to complete the square.

Important Tip:
To solve a quadratic with completing the square method,
You will add and deduct: 1/4 (coefficient of x)2

Subtract 42 to keep the given quadratic unchanged.
x2 + 8x + 4 = 0 is (x2 + 2 × x × 4 + 42) – 42 + 4 = 0,
(x + 4)2 – 12 = 0, i.e.
(x + 4)2 – [√ (12)] 2 = 0, i.e.
(x + 4)2 = [√ (12)] 2, so,
x + 4 = ± √ (12), so,
x + 4 = √ 12, or x + 4 = - √ 12, so
x = 4 + √ 12 or x = 4 – √ 12

Example 2:
Solve the quadratic: 2x2 + x – 4= 0

Solution: You cannot find two numbers whose product is:
(2) × (- 4) i.e. – 8 and whose sum is +1. Use the completing square method.

Use the tips:
2. Add and subtract 1/4 (x coefficient)2

1/2 (2x2 + x – 4) = 0, i.e. x2 + x/2 – 2 = 0
Now add and subtract 1/4 (x coefficient)2 i.e. 1/4 (1/2)2 = 1/4(1/4) = 1/16
The quadratic becomes: x2 + x/2 + 1/161/16 – 2 = 0
Write x/2 as 2 × x × 1/4 to express in complete the square form:
x2 + x/2 + 1/161/16 – 2 = 0,
x2 + 2 × x × 1/4 + 1/16 = 1/16 + 2,
x2 + 2 × x × 1/4 + (1/4)2 = 33/16
(x + 1/4)2 = 33/16 ,
(x + 1/4) = ± √ (33/4)
x + 1/4 = + √ (33/4) or x + 1/2 = - √ (33/4)
x = 1/4 + √ (33/4) or x = 1/4 - √ (33/4)
x = (1 + √ 33)/4 or x = (1- √ 33)/4

Consider the standard quadratic equation: ax2 + bx + c = 0
The roots of a quadratic equation written in standard form are:
x = [–b – √ (b2 – 4ac)]/2a,
x = [–b + √ (b2 – 4ac)]/2a

(We will discuss the proof later on)
1. x2 – 5x + 6 = 0
Solution:
compare the given quadratic with ax2 + bx + c = 0. We see,
a = 1, b = –5, c = 6.
√ (b2 – 4ac)] = √[(-5)2 – 4×1× 6] = √(25 – 24) = √ 1 = 1

Now, the two roots are:
x = [-(-5) – 1]/2 × 1, x = /2 = 2 or
x = [-(-5) + 1]/2× 1, x = /2 = 3

2. 3x2 + 11x – 4= 0
Solution: in this quadratic, a = 3, b = 11, c = - 4
b2 – 4ac = 112 – 4× 3 × (-4) = 121+ 48 = 169
√ (b2 – 4ac)] = √ (169) = 13

Now, the roots are:
x = [-(11) – 13]/2 × 3, x = [-24]/6 = - 4 , or
x = [-(11) +13]/2 × 3, x = /6 = 1/3