Unit 1 Number Systems

Euclid’s Division Lemma

Consider the following folk puzzle.

A trader was moving along a road selling eggs. An idler who didn’t have

much work to do, started to get the trader into a wordy duel. This grew into a

fight, he pulled the basket with eggs and dashed it on the floor. The eggs broke.

The trader requested the Panchayat to ask the idler to pay for the broken eggs.

The Panchayat asked the trader how many eggs were broken. He gave the

following response:

If counted in pairs, one will remain;

If counted in threes, two will remain;

If counted in fours, three will remain;

If counted in fives, four will remain;

If counted in sixes, five will remain;

If counted in sevens, nothing will remain;

My basket cannot accommodate more than 150 eggs.

So, how many eggs were there? Let us try and solve the puzzle. Let the number

of eggs be a. Then working backwards, we see that a is less than or equal to 150:

If counted in sevens, nothing will remain, which translates to a = 7p + 0, for

some natural number p. If counted in sixes, a = 6q+5, for some natural number q.

If counted in fives, four will remain. It translates to a = 5w + 4, for some natural

number w.

If counted in fours, three will remain. It translates to a = 4s + 3, for some natural

number s.

If counted in threes, two will remain. It translates to a = 3t + 2, for some natural

number t.

If counted in pairs, one will remain. It translates to a = 2u + 1, for some natural

number u.

That is, in each case, we have a and a positive integer b (in our example,

b takes values 7, 6, 5, 4, 3 and 2, respectively) which divides a and leaves a remainder

r (in our case, r is 0, 5, 4, 3, 2 and 1, respectively), that is smaller than b. The

* This is modified form of a puzzle given in ‘Numeracy Counts!’ by A. Rampal, and others. moment we write down such equations we are using Euclid’s division lemma,

which is given in Theorem 1.1.

Getting back to our puzzle, do you have any idea how you will solve it? Yes! You

must look for the multiples of 7 which satisfy all the conditions. By trial and error

(using the concept of LCM), you will find he had 119 eggs.

In order to get a feel for what Euclid’s division lemma is, consider the following

pairs of integers:

17, 6; 5, 12; 20, 4

Like we did in the example, we can write the following relations for each such

pair:

17 = 6 × 2 + 5 (6 goes into 17 twice and leaves a remainder 5)

5 = 12 × 0 + 5 (This relation holds since 12 is larger than 5)

20 = 4 × 5 + 0 (Here 4 goes into 20 five-times and leaves no remainder)

That is, for each pair of positive integers a and b, we have found whole numbers

q and r, satisfying the relation:

a = bq + r, 0 £ r < b

Note that q or r can also be zero.

Why don’t you now try finding integers q and r for the following pairs of positive

integers a and b?

(i) 10, 3; (ii) 4, 19; (iii) 81, 3

Did you notice that q and r are unique? These are the only integers satisfying the

conditions a = bq + r, where 0 £ r < b. You may have also realised that this is nothing

but a restatement of the long division process you have been doing all these years, and

that the integers q and r are called the quotient and remainder, respectively.

A formal statement of this result is as follows:

Theorem 1.1 (Euclid’s Division Lemma): Given positive integers a and b,

there exist unique integers q and r satisfying a = bq + r, 0 £ r < b.

This result was perhaps known for a long time, but was first recorded in Book VII

of Euclid’s Elements. Euclid’s division algorithm is based on this lemma.

Euclid’s division algorithm is a technique to compute the Highest Common Factor

(HCF) of two given positive integers. Recall that the HCF of two positive integers a

and b is the largest positive integer d that divides both a and b.

Let us see how the algorithm works, through an example first. Suppose we need

to find the HCF of the integers 455 and 42. We start with the larger integer, that is,

455. Then we use Euclid’s lemma to get

455 = 42 × 10 + 35

Now consider the divisor 42 and the remainder 35, and apply the division lemma

to get

42 = 35 × 1 + 7

Now consider the divisor 35 and the remainder 7, and apply the division lemma

to get

35 = 7 × 5 + 0

Notice that the remainder has become zero, and we cannot proceed any further.

We claim that the HCF of 455 and 42 is the divisor at this stage, i.e., 7. You can easily

verify this by listing all the factors of 455 and 42. Why does this method work? It

works because of the following result.

So, let us state Euclid’s division algorithm clearly.

To obtain the HCF of two positive integers, say c and d, with c > d, follow

the steps below:

Step 1: Apply Euclid’s division lemma, to c and d. So, we find whole numbers, q and

r such that c = dq + r, 0 £ r < d.

Step 2: If r = 0, d is the HCF of c and d. If r ¹ 0, apply the division lemma to d and r.

Step 3: Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.

This algorithm works because HCF (c, d) = HCF (d, r) where the symbol

HCF (c, d) denotes the HCF of c and d, etc.

Example 1: Use Euclid’s algorithm to find the HCF of 4052 and 12576.

Solution:

Step 1: Since 12576 > 4052, we apply the division lemma to 12576 and 4052, to get

12576 = 4052 × 3 + 420

Step 2: Since the remainder 420 ¹ 0, we apply the division lemma to 4052 and 420, to

get

4052 = 420 × 9 + 272

Step 3: We consider the new divisor 420 and the new remainder 272, and apply the division lemma to get

420 = 272 × 1 + 148

We consider the new divisor 272 and the new remainder 148, and apply the division lemma to get

272 = 148 × 1 + 124

We consider the new divisor 148 and the new remainder 124, and apply the division lemma to get

148 = 124 × 1 + 24

We consider the new divisor 124 and the new remainder 24, and apply the division lemma to get

124 = 24 × 5 + 4

We consider the new divisor 24 and the new remainder 4, and apply the division lemma to get

24 = 4 × 6 + 0

The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 4, the HCF of 12576 and 4052 is 4.

Notice that 4 = HCF (24, 4) = HCF (124, 24) = HCF (148, 124) =

HCF (272, 148) = HCF (420, 272) = HCF (4052, 420) = HCF (12576, 4052).

Euclid’s division algorithm is not only useful for calculating the HCF of very

large numbers, but also because it is one of the earliest examples of an algorithm that

a computer had been programmed to carry out.

Remarks:

1. Euclid’s division lemma and algorithm are so closely interlinked that people often

call former as the division algorithm also.

2. Although Euclid’s Division Algorithm is stated for only positive integers, it can be

extended for all integers except zero, i.e., b ¹ 0. However, we shall not discuss this

aspect here.

Euclid’s division lemma/algorithm has several applications related to finding

properties of numbers. We give some examples of these applications below:

Example 2: Show that every positive even integer is of the form 2q, and that every

positive odd integer is of the form 2q + 1, where q is some integer.

Solution: Let a be any positive integer and b = 2. Then, by Euclid’s algorithm,

a = 2q + r, for some integer q ³ 0, and r = 0 or r = 1, because 0 £ r < 2. So,

a = 2q or 2q + 1.

If a is of the form 2q, then a is an even integer. Also, a positive integer can be

either even or odd. Therefore, any positive odd integer is of the form 2q + 1.

Example 3: Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where

q is some integer.

Solution: Let us start with taking a, where a is a positive odd integer. We apply the

division algorithm with a and b = 4.

Since 0 £ r < 4, the possible remainders are 0, 1, 2 and 3.

That is, a can be 4q, or 4q + 1, or 4q + 2, or 4q + 3, where q is the quotient.

However, since a is odd, a cannot be 4q or 4q + 2 (since they are both divisible by 2).

Therefore, any odd integer is of the form 4q + 1 or 4q + 3.

Example 4: A sweet seller has 420 kaju barfis and 130 badam barfis. She wants to

stack them in such a way that each stack has the same number, and they take up the

least area of the tray. What is the number of that can be placed in each stack for this purpose?

Solution: This can be done by trial and error. But to do it systematically, we find

HCF (420, 130). Then this number will give the maximum number of barfis in each

stack and the number of stacks will then be the least. The area of the tray that is used

up will be the least.

Now, let us use Euclid’s algorithm to find their HCF. We have:

420 = 130 × 3 + 30

130 = 30 × 4 + 10

30 = 10 × 3 + 0

So, the HCF of 420 and 130 is 10.

Therefore, the sweet seller can make stacks of 10 for both kinds of barfi.

The Fundamental Theorem of Arithmetic

In your earlier classes, you have seen that any natural number can be written as a

product of its prime factors. For instance, 2 = 2, 4 = 2 × 2, 253 = 11 × 23, and so on.

Now, let us try and look at natural numbers from the other direction. That is, can any

natural number be obtained by multiplying prime numbers? Let us see.

Take any collection of prime numbers, say 2, 3, 7, 11 and 23. If we multiply

some or all of these numbers, allowing them to repeat as many times as we wish,

we can produce a large collection of positive integers (In fact, infinitely many).

Let us list a few:

7 × 11 × 23 = 1771 3 × 7 × 11 × 23 = 5313

2 × 3 × 7 × 11 × 23 = 10626 23 × 3 × 73 = 8232

22 × 3 × 7 × 11 × 23 = 21252

and so on.

Now, let us suppose your collection of primes includes all the possible primes.

What is your guess about the size of this collection? Does it contain only a finite

number of integers, or infinitely many? Infact, there are infinitely many primes. So, if

we combine all these primes in all possible ways, we will get an infinite collection of

numbers, all the primes and all possible products of primes. The question is – can we

produce all the composite numbers this way? What do you think? Do you think that

there may be a composite number which is not the product of powers of primes?

Before we answer this, let us factorise positive integers, that is, do the opposite of

what we have done so far.

We are going to use the factor tree with which you are all familiar. Let us take

some large number, say, 32760, and factorise it as shown:

So, we have factorised 32760 as 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 as a product of

primes, i.e., 32760 = 23 × 32 × 5 × 7 × 13 as a product of powers of primes. Let us try

another number, say, 123456789. This can be written as 32 × 3803 × 3607. Of course,

you have to check that 3803 and 3607 are primes! (Try it out for several other natural

numbers yourself.) This leads us to a conjecture that every composite number can be

written as the product of powers of primes. In fact, this statement is true, and is called the Fundamental Theorem of Arithmetic because of its basic crucial importance to the study of integers. Let us now formally state this theorem.

Theorem 1.2 (Fundamental Theorem of Arithmetic): Every composite number

can be expressed (factorised) as a product of primes, and this factorisation is

unique, apart from the order in which the prime factors occur.

The Fundamental Theorem of Arithmetic says that every composite number

can be factorised as a product of primes. Actually it says more. It says that given

any composite number it can be factorised as a product of prime numbers in a

‘unique’ way, except for the order in which the primes occur. That is, given any

composite number there is one and only one way to write it as a product of primes,

as long as we are not particular about the order in which the primes occur. So, for

example, we regard 2 × 3 × 5 × 7 as the same as 3 × 5 × 7 × 2, or any other

possible order in which these primes are written. This fact is also stated in the

following form:

The prime factorisation of a natural number is unique, except for the order

of its factors.

In general, given a composite number x, we factorise it as x = p1 p 2 ... pn, where

p1

, p2

,..., pn are primes and written in ascending order, i.e., p1

£ p2

£ . . . £ p

n

. If we combine the same primes, we will get powers of primes. For example,

32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 = 23 × 32 × 5 × 7 × 13

Once we have decided that the order will be ascending, then the way the number

is factorised, is unique.

The Fundamental Theorem of Arithmetic has many applications, both within

mathematics and in other fields. Let us look at some examples.

Example 5: Consider the numbers 4n, where n is a natural number. Check whether

there is any value of n for which 4n ends with the digit zero.

Solution: If the number 4n, for any n, were to end with the digit zero, then it would be divisible by 5. That is, the prime factorisation of 4n would contain the prime 5.

not possible because 4n = (2)2n; so the only prime in the factorisation of 4n is 2. So, the

uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no

other primes in the factorisation of 4n. So, there is no natural number n for which 4n

ends with the digit zero.

You have already learnt how to find the HCF and LCM of two positive integers

using the Fundamental Theorem of Arithmetic in earlier classes, without realising it!

This method is also called the prime factorisation method. Let us recall this method

through an example.

Example 6: Find the LCM and HCF of 6 and 20 by the prime factorisation method.

Solution: We have: 6 = 21 × 31 and 20 = 2 × 2 × 5 = 22 × 51.

You can find HCF (6, 20) = 2 and LCM (6, 20) = 2 × 2 × 3 × 5 = 60, as done in your

earlier classes.

Note that HCF (6, 20) = 21 = Product of the smallest power of each common

prime factor in the numbers.

LCM (6, 20) = 22 × 31 × 51 = Product of the greatest power of each prime factor, involved in the numbers.

From the example above, you might have noticed that HCF (6, 20) × LCM (6, 20)

= 6 × 20. In fact, we can verify that for any two positive integers a and b,

HCF (a, b) × LCM (a, b) = a × b. We can use this result to find the LCM of two

positive integers, if we have already found the HCF of the two positive integers.

Example 7: Find the HCF of 96 and 404 by the prime factorisation method. Hence,

find their LCM.

Solution: The prime factorisation of 96 and 404 gives:

96 = 25 × 3, 404 = 22 × 101

Therefore, the HCF of these two integers is 22 = 4.

Also, LCM (96, 404) =

96 404 96 404

9696

HCF (96, 404) 4

Example 8: Find the HCF and LCM of 6, 72 and 120, using the prime factorisation

method.

Solution: We have:

6 = 2 × 3, 72 = 23 × 32, 120 = 23 × 3 × 5

Here, 21 and 31 are the smallest powers of the common factors 2 and 3, respectively.

So, HCF (6, 72, 120) = 21 × 31 = 2 × 3 = 6

23, 32 and 51 are the greatest powers of the prime factors 2, 3 and 5 respectively

involved in the three numbers.

So, LCM (6, 72, 120) = 23 × 32 × 51 = 360

Remark: Notice, 6 × 72 × 120 ¹ HCF (6, 72, 120) × LCM (6, 72, 120). So, the

product of three numbers is not equal to the product of their HCF and LCM.

Revisiting Irrational Numbers
In Class IX, you were introduced to irrational numbers and many of their properties.
You studied about their existence and how the rationals and the irrationals together
made up the real numbers. You even studied how to locate irrationals on the number
line. However, we did not prove that they were irrationals. In this section, we will
prove that 2 , 3 , 5 and, in general, p is irrational, where p is a prime. One of
the theorems, we use in our proof, is the Fundamental Theorem of Arithmetic.
Recall, a number ‘s’ is called irrational if it cannot be written in the form ,
p
q
where p and q are integers and q ¹ 0. Some examples of irrational numbers, with
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which you are already familiar, are :
2 2, 3, 15 , , , 0.10110111011110 . . .
3
p − , etc.
Before we prove that 2 is irrational, we need the following theorem, whose
proof is based on the Fundamental Theorem of Arithmetic.
Theorem 1.3 : Let p be a prime number. If p divides a2, then p divides a, where
a is a positive integer.
*Proof : Let the prime factorisation of a be as follows :
a = p1p2 . . . pn, where p1,p2, . . ., pn are primes, not necessarily distinct.
Therefore, a2 = (p1p2 . . . pn)( p1 p 2 . . . pn) = p2
1p2
2 . . . p2
n.
Now, we are given that p divides a2. Therefore, from the Fundamental Theorem of
Arithmetic, it follows that p is one of the prime factors of a2. However, using the
uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only
prime factors of a2 are p
1, p
2, . . ., p
n. So p is one of p
1, p
2, . . ., p
n.
Now, since a = p1 p2 . . . pn, p divides a.
We are now ready to give a proof that 2 is irrational.
The proof is based on a technique called ‘proof by contradiction’. (This technique is
discussed in some detail in Appendix 1).
Theorem 1.4 : 2 is irrational.
Proof : Let us assume, to the contrary, that 2 is rational.
So, we can find integers r and s (¹ 0) such that 2 =
r
s
.
Suppose r and s have a common factor other than 1. Then, we divide by the common
factor to get 2 ,
a
b
= where a and b are coprime.
So, b 2 = a.
Squaring on both sides and rearranging, we get 2b2 = a2. Therefore, 2 divides a2.
Now, by Theorem 1.3, it follows that 2 divides a.
So, we can write a = 2c for some integer c.
* Not from the examination point of view.
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Substituting for a, we get 2b2 = 4c2, that is, b2 = 2c2.
This means that 2 divides b2, and so 2 divides b (again using Theorem 1.3 with p = 2).
Therefore, a and b have at least 2 as a common factor.
But this contradicts the fact that a and b have no common factors other than 1.
This contradiction has arisen because of our incorrect assumption that 2 is rational.
So, we conclude that 2 is irrational.
Example 9 : Prove that 3 is irrational.
Solution : Let us assume, to the contrary, that 3 is rational.
That is, we can find integers a and b (¹ 0) such that 3 =
a
b
×
Suppose a and b have a common factor other than 1, then we can divide by the
common factor, and assume that a and b are coprime.
So, b 3 = a×
Squaring on both sides, and rearranging, we get 3b2 = a2.
Therefore, a2 is divisible by 3, and by Theorem 1.3, it follows that a is also divisible
by 3.
So, we can write a = 3c for some integer c.
Substituting for a, we get 3b2 = 9c2, that is, b2 = 3c2.
This means that b2 is divisible by 3, and so b is also divisible by 3 (using Theorem 1.3
with p = 3).
Therefore, a and b have at least 3 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that 3 is rational.
So, we conclude that 3 is irrational.
In Class IX, we mentioned that :
l the sum or difference of a rational and an irrational number is irrational and
l the product and quotient of a non-zero rational and irrational number is
irrational.
We prove some particular cases here.
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Example 10 : Show that 5 – 3 is irrational.
Solution : Let us assume, to the contrary, that 5 – 3 is rational.
That is, we can find coprime a and b (b ¹ 0) such that 5 3
a
b
− = ×
Therefore, 5 3
a
b
− = ×
Rearranging this equation, we get
5
3 5 –
a b a
b b
−
= = ×
Since a and b are integers, we get 5 –
a
b
is rational, and so 3 is rational.
But this contradicts the fact that 3 is irrational.
This contradiction has arisen because of our incorrect assumption that 5 – 3 is
rational.
So, we conclude that 5 − 3 is irrational.
Example 11 : Show that 3 2 is irrational.
Solution : Let us assume, to the contrary, that 3 2 is rational.
That is, we can find coprime a and b (b ¹ 0) such that 3 2
a
b
= ×
Rearranging, we get 2
3
a
b
= ×
Since 3, a and b are integers,
3
a
b
is rational, and so 2 is rational.
But this contradicts the fact that 2 is irrational.
So, we conclude that 3 2 is irrational.

Revisiting Rational Numbers and Their Decimal Expansions
In Class IX, you studied that rational numbers have either a terminating decimal
expansion or a non-terminating repeating decimal expansion. In this section, we are
going to consider a rational number, say ( 0)
p
q
q
¹ , and explore exactly when the
decimal expansion of
p
q
is terminating and when it is non-terminating repeating
(or recurring). We do so by considering several examples.
Let us consider the following rational numbers :
(i) 0.375 (ii) 0.104 (iii) 0.0875 (iv) 23.3408.
Now (i)
3
375 375
0.375
1000 10
= = (ii)
3
104 104
0.104
1000 10
= =
(iii)
4
875 875
0.0875
10000 10
= = (iv)
4
233408 233408
23.3408
10000 10
= =
As one would expect, they can all be expressed as rational numbers whose
denominators are powers of 10. Let us try and cancel the common factors between
the numerator and denominator and see what we get :
(i)
3
3 3 3 3
375 3 5 3
0.375
10 2 5 2
×
= = =
×
(ii)
3
3 3 3 3
104 13 2 13
0.104
10 2 5 5
×
= = =
×
(iii)
4 4
875 7
0.0875
10 2 5
= =
×
(iv)
2
4 4
233408 2 7 521
23.3408
10 5
× ×
= =
Do you see any pattern? It appears that, we have converted a real number
whose decimal expansion terminates into a rational number of the form ,
p
q
where p
and q are coprime, and the prime factorisation of the denominator (that is, q) has only
powers of 2, or powers of 5, or both. We should expect the denominator to look like
this, since powers of 10 can only have powers of 2 and 5 as factors.
Even though, we have worked only with a few examples, you can see that any
real number which has a decimal expansion that terminates can be expressed as a
rational number whose denominator is a power of 10. Also the only prime factors of 10
are 2 and 5. So, cancelling out the common factors between the numerator and the
denominator, we find that this real number is a rational number of the form ,
p
q
where
the prime factorisation of q is of the form 2n5m, and n, m are some non-negative integers.
Let us write our result formally:
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Theorem 1.5 : Let x be a rational number whose decimal expansion terminates.
Then x can be expressed in the form ,
p
q
where p and q are coprime, and the
prime factorisation of q is of the form 2n5m, where n, m are non-negative integers.
You are probably wondering what happens the other way round in Theorem 1.5.
That is, if we have a rational number of the form ,
p
q
and the prime factorisation of q
is of the form 2n5m, where n, m are non negative integers, then does
p
q
have a
terminating decimal expansion?
Let us see if there is some obvious reason why this is true. You will surely agree
that any rational number of the form ,
a
b
where b is a power of 10, will have a terminating
decimal expansion. So it seems to make sense to convert a rational number of the
form
p
q
, where q is of the form 2n5m, to an equivalent rational number of the form ,
a
b
where b is a power of 10. Let us go back to our examples above and work backwards.
(i)
3
3 3 3 3
3 3 3 5 375
0.375
8 2 2 5 10
×
= = = =
×
(ii)
3
3 3 3 3
13 13 13 2 104
0.104
125 5 2 5 10
×
= = = =
×
(iii)
3
4 4 4 4
7 7 7 5 875
0.0875
80 2 5 2 5 10
×
= = = =
× ×
(iv)
2 6
4 4 4 4
14588 2 7 521 2 7 521 233408
23.3408
625 5 2 5 10
× × × ×
= = = =
×
So, these examples show us how we can convert a rational number of the form
p
q
, where q is of the form 2n5m, to an equivalent rational number of the form ,
a
b
where b is a power of 10. Therefore, the decimal expansion of such a rational number
terminates. Let us write down our result formally.
Theorem 1.6 : Let x =
p
q
be a rational number, such that the prime factorisation
of q is of the form 2n5m, where n, m are non-negative integers. Then x has a
decimal expansion which terminates.
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We are now ready to move on to the rational numbers
whose decimal expansions are non-terminating and recurring.
Once again, let us look at an example to see what is going on.
We refer to Example 5, Chapter 1, from your Class IX
textbook, namely,
1
7
. Here, remainders are 3, 2, 6, 4, 5, 1, 3,
2, 6, 4, 5, 1, . . . and divisor is 7.
Notice that the denominator here, i.e., 7 is clearly not of
the form 2n5m. Therefore, from Theorems 1.5 and 1.6, we
know that
1
7
will not have a terminating decimal expansion.
Hence, 0 will not show up as a remainder (Why?), and the
remainders will start repeating after a certain stage. So, we
will have a block of digits, namely, 142857, repeating in the
quotient of
1
7
.
What we have seen, in the case of
1
7
, is true for any rational number not covered
by Theorems 1.5 and 1.6. For such numbers we have :
Theorem 1.7 : Let x =
p
q
be a rational number, such that the prime factorisation
of q is not of the form 2n5m, where n, m are non-negative integers. Then, x has a
decimal expansion which is non-terminating repeating (recurring).
From the discussion above, we can conclude that the decimal expansion of
every rational number is either terminating or non-terminating repeating.

Notes on Quadratic Equations

Quadratic Polynomial

A polynomial, whose degree is 2, is called a quadratic polynomial. It is in the form of

p(x) = ax² + bx + c, where a ≠ 0

Quadratic Equation

When we equate the quadratic polynomial to zero then it is called a Quadratic Equation i.e. if

p(x) = 0, then it is known as Quadratic Equation. 

Standard form of Quadratic Equation

where a, b, c are the real numbers and a≠0

Types of Quadratic Equations

1. Complete Quadratic Equation  ax² + bx + c = 0, where a ≠ 0, b ≠ 0, c ≠ 0

2. Pure Quadratic Equation   ax² = 0, where a ≠ 0, b = 0, c = 0

Roots of a Quadratic Equation

Let x = α where α is a real number. If α satisfies the Quadratic Equation ax²+ bx + c = 0 such that aα² + bα + c = 0, then α is the root of the Quadratic Equation.

As quadratic polynomials have degree 2, therefore Quadratic Equations can have two roots. So the zeros of quadratic polynomial p(x) =ax²+bx+c is same as the roots of the Quadratic Equation ax²+ bx + c= 0.

Methods to solve the Quadratic Equations

There are three methods to solve the Quadratic Equations-

1. Factorisation Method

In this method, we factorise the equation into two linear factors and equate each factor to zero to find the roots of the given equation.

Step 1: Given Quadratic Equation in the form of ax² + bx + c = 0.

Step 2: Split the middle term bx as mx + nx so that the sum of m and n is equal to b and the product of m and n is equal to ac.

Step 3: By factorization we get the two linear factors (x + p) and (x + q)

ax² + bx + c = 0 = (x + p) (x + q) = 0

Step 4: Now we have to equate each factor to zero to find the value of x.

These values of x are the two roots of the given Quadratic Equation.

2. Completing the square method

In this method, we convert the equation in the square form (x + a)² - b² = 0 to find the roots.

Step1: Given Quadratic Equation in the standard form ax² + bx + c = 0.

Step 2: Divide both sides by a

Step 3: Transfer the constant on RHS then add square of the half of the coefficient of x i.e.on both sides

Step 4: Now write LHS as perfect square and simplify the RHS.

Step 5: Take the square root on both the sides.

Step 6: Now shift all the constant terms to the RHS and we can calculate the value of x as there is no variable at the RHS.

3. Quadratic formula method

In this method, we can find the roots by using quadratic formula. The quadratic formula is

where a, b and c are the real numbers and b² – 4ac is called discriminant.

To find the roots of the equation, put the value of a, b and c in the quadratic formula.

Nature of Roots

From the quadratic formula, we can see that the two roots of the Quadratic Equation are -

Where D = b² – 4ac

The nature of the roots of the equation depends upon the value of D, so it is called the discriminant.

∆ = Discriminant

 

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