In this section, we will take up the Quadratic Equations part and deal with all the concepts and formulae in it. We will also see some Cubic Equations formulae and solve few problems on them  

• Quadratic Equations are equations of degree 2 i.e. the power of at least one variable must be 2 and also the maximum power on any variable should not exceed 2.

It is written as ax² + bx + c = 0, where a and b are coefficients of x² and x respectively and c is a constant.

For example, x² + 8x + 4 = 0 is a quadratic equation.

• Methods to solve Quadratic Equations

Quadratics equations can be solved by two methods:

1. Factorisation Method.
2. Sridharacharya’s Method.

• By Factorisation Method

If we can factorise ax2 + bx + c as (x – α) (x – β), then ax2 + bx + c = 0 is equivalent to (x – α) (x – β) = 0.

Therefore,

(x – α) (x – β) = 0

⇒ (x – α) = 0 or (x – β) = 0

⇒ x = α or x = β

We call α and β are called roots of equation ax² + bx + c = 0.

Let’s take few examples and see what type of questions come in it and how do we use factorization method.

E.g. 1: Factors of x² + 7x + 10 are?

Step 1: Split the middle term in two terms such that its coefficient value doesn’t change and the product of the terms is equal to the products of first and third term.

Here, we have to break 7x in two terms such that the coefficient remains 7 but the product is 10, i.e. (a + b) = 7 and ab = 10.

So, we get a = 5 and b = 2. This fulfills both the criteria.

Putting in equation,

x2 + 7x + 10

⇒ x2 + 5x + 2x + 10

Step 2: Separate the terms which have anything common in two different brackets.

⇒ x2 + 5x + 2x + 10

⇒ (x2 + 5x) + (2x + 10)

⇒ x(x + 5) + 2(x + 5)

Step 3: Again separate the common terms.

Here, we see that (x + 5) is common in both, separating it.

⇒ x(x + 5) + 2(x + 5)

⇒ (x + 5)(x + 2) (Ans).

E.g. 2: Find the factors of x² – x – 132.

x2 – x – 132

Here, we have to split middle term such that a + b = -1 and ab = -132.

So, we have a = -12 and b = 11

x2 – x – 132

⇒ x2 – 12x + 11x – 132

⇒ x(x-12) + 11(x-12)

⇒ (x – 12)(x + 11) (Ans).

• By Sridharacharya’s Method

If we have quadratic equation in the form ax² + bx + c = 0, then the roots of the equation are:

α = (-b + √D)/2a and β = (-b – √D)/2a

Where d is called Discriminant of quadratic equation, D = b² – 4ac

Factors are: (x – α) (x – β)

We’ll solve few examples from NTPC Railway Exams to understand it in a better way.

• E.g. 1: Factors of x² + 7x + 10 are?

Step 1: Identify a, b and c. Find d.

x2 + 7x + 10

Here, a = 1, b = 7 and c = 10

Discriminant, D = b2 + – 4ac

                           = 72 – 4 × 1 × 10

                           = 49 – 40

                           = 9

Step 2: Put the values in formula.

α = (-b + √D)/2a

⇒ α = (-7 + √9)/(2 × 1)

⇒ α = (-7 + 3)/2

⇒ α = -4/2

⇒ α = -2

And,

β = (-b – √D)/2a

⇒ β = (-7 – √9)/(2 × 1)

⇒ β = (-7 – 3)/2

⇒ β = -10/2

⇒ β = -5

Step 3: Find the factors by putting value of α and β

Hence, Required factors = (x – α)(x – β)

                                     = {x – (-2)} {x – (-5)}

                                     = (x + 2)(x + 5) (Ans).

E.g. 2: Find the factors of x² + 8x + 12 are

Here, a = 1, b = 8, c = 12

D = b2 – 4ac

⇒ D = 82 – 4 × 1 ×12

    = 64 – 48

    = 16

Putting in formula,

α = (-8 + √16)/(2 × 1)

⇒ α = (-8 + 4)/2

⇒ α = -4/2

⇒ α = -2

And,

β = (-8 – √16)/(2 × 1)

⇒ β = (-8 – 4)/2

⇒ β = -12/2

⇒ β = -6

Hence, Factors = (x – α)(x – β)

                       = {x – (-2)} {x – (-6)}

                       = (x + 2)(x + 6) (Ans).

• Basic formulae for Quadratic Equations

We have some basic formulae that help us to simplify the arithmetical expressions. We need to keep them on our fingertips to get quick and correct results.

There have been direct questions based on these formulae.

1. (a + b)2 = a2 + 2ab + b2
2. (a – b)2 = a2 – 2ab + b2
3. (a + b)2 + (a – b)2 = 2(a2 + b2)
4. (a + b)2 – (a – b)2 = 4ab
5. a2 – b2 = (a + b)(a – b)
6. (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

Let’s take problems questions and solve them to understand the application of these formulae.

• E.g. 1: Expand: (w – 9)²

Without formula,

(w – 9)2 = (w – 9)(w – 9) (Multiplying each term with each other)

           = w2 – 9w – 9w + 81

           = w2 – 18w + 81 (Ans).

Using formula,

(w – 9)2 = w2 – 2×w×9 + 92        [(a – b)2 = a2 – 2ab + b2]

           = w2 – 18w + 81 (Ans). 

E.g. 2: Simplify: (81x² – 49y²) / (9x + 7y)

We have five formulas in square form. Try to convert the equation in any one of them. Here, it is clearly visible that that 81 and 49 are squares. So, we can convert them in a² – b² form.

(81×2 – 49y2)/(9x + 7y)

⇒ (92 x2 – 72y2)/(9x + 7y)

⇒ {(9x)2 – (7y)2}/(9x + 7y)                      [∵ a2 – b2 = (a + b)(a – b)]

⇒ (9x + 7y)(9x – 7y)/(9x + 7y)                [Cancelling out (9x + 7y)]

⇒ (9x – 7y) (Ans).

E.g. 3: If (a + b + c) = 6 and a² + b² + c² = 14, then (ab + bc + ca) =?

We have,

(a + b + c) = 6

We know the formula for (a + b + c)2.

Therefore, squaring on both sides

⇒ (a + b + c)2 = 62

⇒ a2 + b2 + c2 + 2ab + 2bc + 2ca = 36

Putting the given value of a2 + b2 + c2

⇒ 14 + 2ab + 2bc + 2ca = 36

⇒ 2(ab + bc + ca) = 36 – 14

⇒ (ab + bc + ca) = 22/2

⇒ (ab + bc + ca) = 11 (Ans).

Cubic Equations are equations of degree 3 i.e. the maximum power on any variable is 3.

• Basic Formulae on Cubic Equations
• (a + b)3 = a3 + b3 + 3ab (a + b)
• (a – b)3 = a3 – b3 – 3ab (a – b)
• a3 + b3 = (a + b)( a2 – ab + b2)
• a3 –  b3 = (a – b)( a2 + ab + b2)

E.g. 1: Expand: (s + 2)³

Without formula,

(s + 2)3 = (s + 2)(s + 2)(s + 2)

          = (s2 + 2s + 2s + 4)(s + 2)

          = (s2 + 4s + 4)(s + 2)

          = s3 + 2s2 + 4s2 + 8s + 4s + 8

          = s3 + 6s2 + 12s + 8

Using formula of (a + b)3,

(s + 2)3 = s3 + 23 + 3 × s × 2 (s + 2)

           = s3 + 8 + 6 × s (s + 2)

           = s3 + 6s2 + 12s + 8 (Ans). 

E.g. 2: Expand (a – 4)³

By formula,

(a – b)³ = a³ – b³ – 3ab (a – b)

(a – 4)³ = a³ – 4³ – 3 × a × 4 (a – 4)

            = a³ – 64 – 12a × (a – 4)

            = a³ – 64 – 12a × a – 12a × (-4)

            = a³ – 64 – 12a² + 48a

            = a³ – 12a² + 48a – 64 (Ans).