# [Maths Class Notes] on Quadratics Pdf for Exam

### Quadratic Equation: Definition

The simple definition of a quadratic equation is the polynomial equation whose highest order is two. It commonly gets expressed as ax² + bx + c = 0. In which, x is the unknown variable, and a, b, c are the constant terms. Also, note that ‘a’ is never equal to 0; otherwise, the equation becomes a linear one.

Since the quadratic has a single unknown term or variable, it also gets referred to as a univariate. The power of the variable x always has to be a non-negative integer. Thus, it becomes a polynomial equation with the highest degree as two.

The solution for the equation is the values of x, and they get called as zeros. They are the final solutions which satisfy the equation. When it comes to quadratics, there are two zeros or roots of the equation. When you insert the value of x in the L.H.S. of the equation, then you get a zero. That’s why they get called as zero.

There are two fundamental concepts to solve a quadratic equation: 1. Formula method, and 2. Factorization method. These are the quickest methods to solve any quadratic equation example. You can learn both of these methods, as follows.

### Solving Quadratics by Formula

By learning the quadratic equation formula, you can solve any quadratic equation quickly. If the quadratic equation looks like ax² + bx + c = 0, then below is the formula you need to apply.

The signs (+/–) in the formula indicate that you obtain two values/solutions for the x.

### Examples of the Quadratic Equations :

Below are the examples for quadratic equations of the form: ax² + bx + c = 0.

• x² –x – 7 = 0

• 4x² – 2x – 9 = 0

• –x² + 2x + = 0

Below are the examples for quadratic equations where ‘c’ or a constant term is absent.

• -x² – 6x = 0

• x² + 4x = 0

• -14x² + 9x = 0

Below are the examples for quadratic equations where ‘bx’ or a linear coefficient is absent.

• x² – 14 = 0

• 5x² + 54 = 0

• -x² – 7 = 0

Below are the examples for quadratic equations in factored form.

• (x – 6)(x + 1) = 0 (after solving, you get x² – 5x – 6 = 0)

• (2x+3)(3x – 2) = 0 (after solving, you get 6x² + 5x – 60)

• (x – 4)(x + 2) = 0 (after solving, you get x² – 2x – 8 = 0)

### Solving Quadratics by Factoring

Apart from the quadratic equation formula, factorization is another method of obtaining solutions for quadratic equations. Below are steps to find the solution of the quadratics by factoring.

• You begin with an equation in the form of ax² + bx + c = 0.

• Then, you factor the L.H.S. of the equation while assuming zero on the R.H.S. of the equation.

• By assigning each factor to zero, you can solve the equation to obtain the values of x.

When the main coefficient is not equal to zero, then have to arrange the factors in a way as below.

Consider an equation: 2x² – x – 6 = 0

(2x + 3) (x – 2) = 0

2x + 3 = 0

X = -[frac{3}{2}]

In the end, you get: X = 2.

### Solved Examples

Using quadratic equations, you can solve word problems, typical equations, which involve determining the speed, area, etc. Below you can find solved quadratic equation example to help you understand the topic even better.

Question 1: Find the value of x: 27x2 − 12 = 0

A) 2/3 B) ± 2/3 C) Ambiguous

Answer : Here, a = 27, b = 0 and c = -12.

Now, by putting the values in the quadratic equation formula, you get:

x = [frac{-0 pm sqrt{0^{2} – 4(27)(-12)}}{2(27)}]

x = ± [sqrt{frac{4}{9}}]

Finally, x = ± [frac{2}{3}]. So, the correct option is B.

Question 2: The area of a rectangle is 336 cm². Its length is four more than twice its width. Find its actual width.

Answer: Consider the width of a rectangle as ‘x.’

From the given data, length = (2x + 4) cm

As you know, Area of rectangle = Length x Width

Now, we get the equation as x(2x + 4) = 336

By further solving, 2x² + 4x – 336 = 0

x² + 2x – 168 = 0

x² + 14x – 12x – 168 = 0

x (x + 14) – 12 (x + 14) = 0

(x + 14) (x – 12) = 0

x = -14, x = 12

Since a measurement cannot be negative; the width of the rectangle is, x = 12 cm.