CHAPTER – 2 : Polynomials

DEGREE OF A POLYNOMIAL:-

If p(x) is a polynomial in x, the highest power of x in p(x) is called the degree of polynomial p(x).

And expressions like:- x2−−√+2,1x−1etc., are not polynomials.

Example-1:

3x⁴+2x²+6x = 1, the highest power of the given polynomial equation is 4

Therefore, its degree is 4.

Example-2:

x⁵y³z + 2xy³+4x²yz²

This equation is of multiple variables (x, y, z) and to find the degree of this equation, we just need to add up the degrees of the variables in each of the terms and also it does not matter whether they are different variables.

Therefore, 1^st term (5+3+1), 2^nd term (1+3) and 3^rd term (2+1+2).

The highest total is 9, of the 1^st term,

Therefore, its degree is 9.

Polynomial equation with degree 1 is called a linear polynomial.

Example:     x – 3–√ = 0

Polynomial equation with degree 2 is called a quadratic polynomial.

Example:    x2+2–√x+7=0.

Polynomial equation with degree 3 is called a cubic polynomial.

Example:    3x3–5x2+7–√x+5–√=0.

Factorization of a quadratic polynomial equation by splitting middle term:-

General form of quadratic polynomial equation:- ax² + bx + c = 0

Step 1:- Find the product of a (first term) and c (last term).

Step 2:- Then split the product of a and c into two numbers such that their product remains same as (a × c) and their sum or difference is equal to the value of b (middle term).

Step 3:- Now, there will be total four terms in the equation, make two groups with each group having two terms each. Then take whatever is common and equate it to 0, the resultant values of x will be factors of the quadratic equation.

Example:

12 x² = -11x +15

Solution:

12x ² + 11x -15 = 0                (product of a and c is 12×15=180)

12x ² + 20x – 9x -15 = 0      (splitting 180 into 20 and 9 such that, 20 – 9 = 11 and 20×9 = 180).

4x(3x + 5) – 3(3x + 5) = 0

(4x -3)(3x + 5) = 0

4x – 3 = 0 or 3x + 5 = 0

4x = 3 or 3x = – 5

x = 34 or x = −53

Therefore, solution of this quadratic equation  is (−53, 34 ).

This solution is called zeroes or roots of the given polynomial.

                                                                          EXERCISE 2.1

Graphical method to find zeroes:-

Total number of zeroes in any polynomial equation = total number of times the curve intersects x-axis.

Example 1:-

In the above graph, the curve intersects x axis at A and B.

Therefore, the equation of the curve will have maximum of 2 zeroes.

Example 2:-

From the above graph, the curve cuts the x-axis at point A

Therefore, number of zeroes of p(x) = 1

Example 3: –

From the above graph, the curve p(x) intersects x-axis at points A, B and C.

Therefore, the number of zeroes of polynomial equation is 3.

Zeroes and Coefficient of a polynomial: 

If α and β are zeroes of any quadratic polynomial ax² + bx + c=0,

Then, Sum of zeroes = α+β = −ba

Product of zeroes = αβ = ca

If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as: 

x² – (α + β)x + αβ = 0                                    

                                    EXERCISE 2.2

Q.1 Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) 6x² + 11x + 5 = 0

Sol_n: –

6x² + 11x + 5 = 6x² + 6x + 5x + 5

= 6x(x +1) + 5(x +1)

= (x +1) (6x +5)

∴ zeroes of polynomial equation 6x² +11x +5 are { −1, −56 }

Now, Sum of zeroes of this given polynomial equation = −1+( −56 ) = −116

But, the Sum of zeroes of any quadratic polynomial equation is given by = −coeff.ofxcoeff.ofx2

= −116

And Product of these zeroes will be = −1×−56 = 56

But, the Product of zeroes of any quadratic polynomial equation is given by = constanttermcoeff.ofx2

= 56

Hence the relationship is verified.

(ii) 4s² – 4s + 1

Sol:

     4s² – 4s + 1 = 4s² – 2s – 2s + 1

= 2s (2s – 1) –1(2s – 1)

= (2s – 1) (2s – 1)

∴ zeroes of the given polynomial are: {12,12}

∴ Sum of these zeroes will be =  = 1.

But, The Sum of zeroes of any quadratic polynomial equation is given by = −coeff.ofscoeff.ofs2

= −44 = 1

And the Product of these zeroes will be = 12×12
=14

But, Product of zeroes in any quadratic polynomial equation is given by = constanttermcoeff.ofs2

= 14.

Hence, the relationship is verified.

(iii) 6x² – 3 – 7x

Sol:

     6x² – 7x – 3 = 6x² – 9x + 2x – 3

= 3x (2x – 3) +1(2x – 3)

= (3x + 1) (2x – 3)

∴ zeroes of the given polynomial are: – (−13,32)

∴ sum of these zeroes will be =  −13+32
=76

But, The Sum of zeroes in any quadratic polynomial equation is given by = −coeff.ofxcoeff.ofx2

= 76

And Product of these zeroes will be = −13×32=−12
Also, the Product of zeroes in any quadratic polynomial equation is given by = constanttermcoeff.ofx2

= −36 = −12

Hence, the relationship is verified.

(iv) 4u² + 8u

Sol:

     4u² + 8u = 4u (u+2)

Clearly, for finding the zeroes of the above quadratic polynomial equation either: – 4u=0 or u+2=0

Hence, the zeroes of the above polynomial equation will be (0, −2)

∴ Sum of these zeroes will be = −2

But, the Sum of the zeroes in any quadratic polynomial equation is given by = −coeff.ofucoeff.ofu2

 = −84 = −2

And product of these zeroes will be = 0 × −2 = 0

But, the product of zeroes in any quadratic polynomial equation is given by = constanttermcoeff.ofu2                                                                          = −04 = 0

Hence, the relationship is verified.

(v) t² – 15

Sol:

     t² – 15 = (t+ 15−−√) (t − 15−−√)

Therefore, zeroes of the given polynomial are: – {15−−√, −15−−√}

∴ sum of these zeroes will be = 15−−√ −  15−−√ = 0

But, the Sum of zeroes in any quadratic polynomial equation is given by = −coeff.ofxcoeff.ofx2

                                                                = −01 = 0   

And the product of these zeroes will be = (15−−√) × (−15−−√)  = −15

But, the product of zeroes in any quadratic polynomial equation is given by

= constanttermcoeff.oft2                                                                     = −151  = −15

Hence, the relationship is verified.

(vi) 3x² – x – 4

Sol:

 3x² − x − 4   = 3x² – 4x + 3x − 4

= x (3x – 4) +1(3x – 4)

= ( x + 1) (3x – 4)

∴ zeroes of the given polynomial are: – {−1, 43 }

∴ sum of these zeroes will be = −1 +43 = 13

But, the Sum of zeroes in any quadratic polynomial equation is given by = −coeff.ofxcoeff.ofx2

= −(−1)3 = 13

And the Product of these zeroes will be = {−1 × 43 }

= −43

But, the Product of zeroes in any quadratic polynomial equation is given by = constanttermcoeff.ofx2

= −43

Hence, the relationship is verified.

Q2. Form a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i). 26 , −3

Sol.

Given,

α + β = 26

αβ = −3

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x² – (α+β)x +αβ = 0

Thus, the required quadratic equation will be:

x² – (26)x −3 = 0

6x² − 2x – 18 = 0.

(ii). 3–√ , 43

Sol.

Given,

α + β = 3–√

αβ = 43

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as: –

x² – (α+β)x +αβ=0

Thus, the required quadratic equation will be: –

x² – (3–√)x + 43 =0

3x² − 33–√x + 4 = 0.

(iii).  0, 7–√

Sol.

Given,

α + β = 0

αβ = 7–√

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x² – (α+β)x +αβ=0

Thus, the required quadratic equation will be: –

x² – (0)x + 7–√ = 0

x² + 7–√ = 0.

(iv).  −2, −2

Sol.

Given,

α + β = −2

αβ = −2

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:

x² – (α+β)x +αβ=0

∴ The required quadratic polynomial will be:

x² – (−2)x −2 = 0

x² + 2x – 2 = 0.

(v). −72, 39

Sol.

Given,

α + β = −72

αβ = 39

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as: –

x² – (α+β)x +αβ = 0

∴ The required quadratic polynomial will be:

x² – (−72)x + 39 = 0

18x² + 63x + 6 = 0.

(vi). 6, 0

Sol.

Given,

α + β = 6

αβ = 0

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x² – (α+β)x +αβ = 0

∴ The required quadratic polynomial will be:-

x² – 6x + 0 = 0

x² – 6x = 0.

                                  EXTRA QUESTIONS

Q.1 Find a quadratic polynomial whose zeroes are: –2+12√, 2−12√.

Sol.

Given: –

α + β = 2+12√ + 2−12√

= 4.

αβ = (2+12√)(2−12√)=4−12=72

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as: –

x² – (α+β)x +αβ = 0

Thus, the required quadratic polynomial equation will be :-

x² – (4)x −72 = 0

2x²−8x+7 = 0.

Q.2 If α and β are the roots of a quadratic polynomial ax²+bx+c, then find the value of  α² + β².

Sol.

From the equation, (α+β=−ba)

And,               α×β=ca

(α+β)2=α2+β2+(2αβ)

∴ α2+β2=(α+β)2−2αβ

∴ α2+β2=(−ba)2−2ca

∴ α2+β2=b2a2−2ca

 ∴α2+β2=b2−2aca2.

Similarly, we can find out the values of (α³ + β³) and (α³ – β³).

Q3. If α and β are zeroes of a quadratic polynomial x²+4x+3, form the polynomial whose zeroes are  1+αβand1+βα.

Sol.

Since α and β are zeroes of a quadratic polynomial x²+4x+3,

α+β= -4, αβ = 3

Given: –  α₁ = 1+αβ

β₁ = 1−βα

Now, sum of zeroes = 1+αβ+1+βα

= 2+αβ+βα

= 2αβ+α2+β2αβ

= (α+β)2αβ

On putting values of α+β and αβ from above we get:-

Sum of zeroes = α₁ + β₁ = −423

= 163

Now, Product of zeroes = (1+αβ)(1+βα)

=1+βα+αβ+αββα

=2αβ+β2+α2αβ

= (α+β)2αβ

On putting values of α+β and αβ we get:

Product of zeroes = α₁ × β₁ = −423

= 163

Thus the required quadratic polynomial equation will be:-

x² – (α₁ + β₁)x + α₁β₁ = 0

x² – (163)x + 163 = 0

3x² – 16x +16=0.

Q4. If α and β are zeroes of a quadratic polynomial p(x) = rx²+4x+4, Find the values of “r” if: – α² + β² = 24.

Sol.

From the given polynomial p(x),

α + β = −4r, and αβ = −4r………. (1)

Since,(α + β)² = α² + β² + 2αβ

Given,   α² + β² = 24 and from equation (1).

Therefore,  (−4r)2=24+2×4r

16r2=24+2×4r

16 = 24 r² + 8r

3r² + r – 2 = 0

3r² + 3r – 2r -2 = 0

3r(r+1) – 2(r+1) = 0

(3r-2) (r+1) = 0

∴  r = 23   or   r = -1

                                 DIVISION ALOGORITHM

If suppose p(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that: –

p(x)=g(x) × q(x) + r(x), where r(x)=0 or  degree of r(x) < degree of g(x) …….(1)

This is known as the division algorithm for polynomials.

And since, Dividend = Divisor × Quotient + Remainder

On comparing it with equation (1) we can conclude that:

Dividend = p(x).

Divisor = g(x).

Quotient = q(x).

Remainder = r(x).

                              EXERCISE 2.3

Q.1 If a polynomial x³ -3x²+x+2 is divided by a polynomial g(x), the quotient and remainder obtained are (x-2) and (-2x+4), respectively. Find the equation of g(x).

Sol.

Since, Dividend = Divisor × Quotient + Remainder

Therefore, x³ -3x²+x+2 = g(x) × (x-2) + (-2x+4)

(x³ -3x²+x+2) – (-2x+4) = g(x) × (x-2)

Therefore, g(x) × (x-2) = x³ -3x²+3x-2

Now, for finding g(x) we will divide “x³ -3x²+3x-2” with (x-2)

Therefore, g(x) = (x² – x +1)

Q.2 Find the quotient and remainder by dividing the polynomial f(x) by the polynomial g(x).

(i) f(x) = x³ + 2x² – 9x + 5,   g(x) = x²+5

Therefore, Quotient is (x+2) and Remainder is (-14x − 5)

(ii) f(x) = x⁵+2x⁴-9x³+5x²-2x+1, g(x) = x³ + x² – x+1

Therefore, Quotient is (x² + x − 9) and Remainder is (14x² – 12x +10).

(iii) f(x) = 2x⁴ + 7x³ + 5x² + 8x + 5, g(x) = 11− 2x³ + x²

Therefore, Quotient is  – (x + 4) and Remainder is (9x² + 19x +49).

Q3. Find all the zeroes of the polynomial equation 2x⁴-3x³-3x²+6x-2, if two of its zeroes are 2–√and−2–√.

Sol.

Since this is a polynomial equation of degree 4, there will be a total of 4 roots.

Let, f(x) = 2x⁴ – 3x³ – 3x² + 6x – 2

2–√and−2–√ are zeroes of f(x).

∴(x−2–√)(x+2–√)= (x²− 2) = g(x), is a factor of given polynomial f(x).

If we divide f(x) by g(x), the quotient will also be a factor of f(x) with remainder =0.

So, 2x⁴ − 3x³ − 3x² + 6x – 2 = (x² – 2) (2x² – 3x +1).

Now, on further factorizing (2x² – 3x +1) we get,

2x² – 3x +1 = 2x² – 2x − x +1 = 0

2x (x − 1) – 1(x−1) = 0

(2x−1) (x−1) = 0

So, its zeroes are given by:  x= 12 and x = 1

Therefore, all four zeroes of the given polynomial equation are:

2–√,−2–√,12and1.

Q4. Find all zeroes of a polynomial equation x⁴-6x³-26x²+138x-35, if two of its zeroes are 2+3–√ and 2−3–√.

Sol.

Since this is a polynomial equation of degree 4, hence there will be total 4 roots.

Let, f(x) = x⁴ – 6x³ – 26x²+ 138x – 35

Since, (2+3–√)and(2−3–√) are zeroes of given polynomial f(x).

∴[x−(2+3–√)][x−(2−3–√)]=0

(x−2−3–√)(x−2+3–√)

On multiplying the above equation we get,

x² −4x + 1, this is a factor of a given polynomial f(x).

Now, if we will divide f(x) by g(x), the quotient will also be a factor of f(x) and the remainder will be 0.

So, x⁴ − 6x³  − 26x² + 138x – 35 = (x² – 4x + 1) (x² – 2x −35).

Now, on further factorizing (x² – 2x −35) we get,

x² – (7−5)x − 35 = x² – 7x + 5x +35 = 0

x(x − 7) + 5 (x−7) = 0

                                         (x+5) (x−7) = 0

So, its zeroes are given by:              x= −5 and x = 7.

Therefore, all four zeroes of given polynomial equation are:

2+3–√,2−3–√, −5 and 7.

Q5. Find all zeroes of a polynomial equation 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are 53−−√ and −53−−√.

Sol.

Since this is a polynomial equation of degree 4, hence there will be total 4 roots.

Let, f(x)= 3x⁴+6x³-2x²-10x-5

53−−√ and −53−−√ are zeroes of polynomial f(x).

∴(x−53−−√)(x+53−−√)

 (x2−53)=0

 (3x²−5)=0, is a factor of given polynomial f(x).

Now, when we will divide f(x) by (3x²−5) the quotient obtained will also be factor of f(x) and remainder will be 0.

Therefore, 3x⁴ + 6x³ − 2x² − 10x – 5 = (3x² – 5) (x² + 2x +1).

Now, on further factorizing (x² + 2x +1) we get,

x² + 2x +1 = x² + x + x +1 = 0

x(x + 1) + 1(x+1) = 0

                     (x+1) (x+1) = 0

So, its zeroes are given by: x= −1 and x = −1

Therefore, all four zeroes of given polynomial equation are:

53−−√, −53−−√, −1 and −1.

Related Links

Chapter-1 : Real Numbers

QUESTION-1

Use Euclid’s division algorithm to find the HCF of:

i)135 and 225

ii)196 and 38220

iii)867 and 225

Solution:

i) We start with the larger number i.e 225

By Euclid’s division algorithm,we have

225=1×135+90

135=1×90+45

90=2×45+0

Hence, HCF(225,135)=HCF(135,90)=HCF(90,45)=45

Therefore,the HCF of 135 and 225 is 45

ii) We start with the larger number i.e 38220

By Euclid’s division algorithm,we have

38220=196×195+0

196=196×1+0

Hence , HCF(196,38220)=196

Therefore,the HCF of 196 and 38220 is 196

iii) We start with the larger number i.e 867

By Euclid’s division algorithm,we have

867=225×3+102

225=102×2+51

102=51×2+0

Hence, HCF(867,225)=HCF(225,102)=HCF(102,51)=51

Therefore,the HCF of   867 and 225 is 51

Question 2:

Show that any positive odd integer is of the form 6q+1,6q+3 & 6q+5

where q is some integer.

Solution :

Using Euclid’s division algorithm,we have

x=bq+r {0≤r<b}…..(1)

Substituting b=6in equation(1)

So, x=6q+r,where r=0,1,2,3,4,5

If r=0, x=6q+0(divisible by 2)…..even

r=1, x=6q+1(not divisible by 2)…..odd

r=2, x=6q+2(divisible by 2)…..even

r=3, x=6q+3(not divisible by 2)…..odd

r=4, x=6q+4(divisible by 2)…..even

r=5, x=6q+5(not divisible by 2)…..odd

Therefore,the number 6q,6q+1,6q+2,6q+3,6q+4,6q+5are either even or odd.Hence ,any positive odd integer is of the form 6q+1,6q+3 & 6q+5Where q is some integer.

Question 3:

An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

Solution :

The maximum number of columns would be the HCF of (616,32)

We can find the HCF of 616 and 32 by using Euclid Division Algorithm.

Therefore,

616=19×32+8

32=4×8+0

8=8×1+0

So , HCF of 6161 and 32 is 8

Hence ,the maximum number of columns in which they can march is 8.

Question 4:

Use Euclid’s Division lemma to show that the square of any positive integer is either of the form 3m, 3m+1 for some integer m.

Solution :

According to Euclid Algorithm ,

We have x=bq+r …..(1)

Substituting , b=3

We get,x=3q+r (where,0≤r<3

i.e r=0,1,2 )

When ,r=0 ,x=3q…….(A)

r=1 ,x=3q+1…(B)

r=2 ,x=3q+2…(C)

So, Squaring eqn(A),(B) and (C)

We get,

From eqn (A),

x2=9q2 x2=3×3q2

x2=3×m (where,m=3q2)

From eqn (B),

x2=(3q+1)2

=9q2+1+6q

=9q2+6q+1

=3(3q2+2q)+1

=3m+1 (where,m=3q2+2q)

From eqn (C),

x2=(3q+2)2

=9q2+4+12q

=9q2+12q+3+1

=3(3q2+4q+1)+1

=3m+1 (where,m=3q2+4q+1)

Hence, any positive integer is either of the form 3m,3m+1 for some integer m.

EXERCISE 1.2

Question 6:

Express each number as a product of its prime factors:

i)140

ii)156

iii)3825

iv)5005

v)7429

Solution :

i) 140

Take  LCM of 140 i.e  2×2×5×7×1

Hence,  140=2×2×5×7×1

ii) 156

Take LCM of 156 i.e 2×2×13×3×1

Hence, 156=2×2×13×3×1

iii)3825

Take  LCM of 3825 i.e 3×3×5×5×17×1

Hence, 3825=3×3×5×5×17×1

iv)5005

Take LCM of 5005 i.e 5×7×11×13×1

Hence, 5005=5×7×11×13×1

v)7429

Take LCM of 7429 i.e 17×19×23×1

Hence, 7429=17×19×23×1

Question 7:

Find the LCM and HCF of the following pairs of integer and verify that LCM×HCF=Product of the two numbers.

i) 26 and 91

ii) 510 and 92

iii) 336 and 54

Solution :

i) 26 and 91

26=2×13×1(expressing as product of it’s prime factors)

91=7×13×1(expressing as product of it’s prime factors)

So, LCM(26,91)=2×7×13×1=182

HCF(26,91)=13×1=13

Verification:

LCM×HCF=13×182=2366

Product of 26 and 91 =2366

Therefore,LCM×HCF=Product of the two numbers .

i) 510 and 92

510=2×3×17×5×1(expressing as product of it’s prime factors)

92=2×2×23×1(expressing as product of it’s prime factors)

So,

LCM(510,92)=2×2×3×5×17×23=23,460

HCF(510,92)=2

Verification:

LCM×HCF=23,460×2=46,920

Product of 510 and 92 =46,920

Therefore,LCM×HCF=Product of the two numbers .

iii) 336 and 54

336=2×2×2×2×7×3×1(expressing as product of it’s prime factors)

54=2×3×3×3×1(expressing as product of it’s prime factors)

So,

LCM(336,54)=24×33×7=3024

HCF(336,54)=2×3=6

Verification:

LCM×HCF=3024×6=18,144

Product of 336 and 54=18,144

Therefore,LCM×HCF=Product of the two numbers .

Question 8:

Find the LCM and HCF of the following integers by applying the prime factorization method.

i) 12,15 and 21

ii) 17,23 and 29

iii) 8,9 and 25

Solution :

i) 12,15 and 21

12=2×2×3

15=5×3

21=7×3

From the above ,HCF(12,15,21)=3and LCM(12,15,21)=420

ii)17,23,and 29

17=17×1

23=23×1

29=29×1

From the above ,HCF(17,23,29)=1and LCM(17,23,29)=11339

iii)8,9 and 25

8=2×2×2

9=3×3

25=5×5

From the above ,HCF(8,9,25)=1and LCM(8,9,25)=1800

Question 9:

Given that HCF(306,657)=9 .Find LCM(306,657)?

Solution :

We know,

HCF×LCM=Product of two numbers

i.e 9×LCM=306×657

LCM=306×6579=22338

Question 10:

Check whether 6n can end with the digit 0 for any natural number n.

Solution :

If the number 6n ends with the digit zero,then it is divisible by 5.Therefore the prime factorization of 6n contains the prime 5.This is not possible because the only prime in the factorization of 6n is 2 and 3 and the uniqueness of the fundamental theorem of arithmetic guarantees that these are no other prime in the factorization of 6n

Hence, it is very clear that there is no value of n in natural number for which 6n ends with the digit zero.

Question 11:

Explain why 7×11×13+13 and 7×6×5×4×3×2×1+5 are composite number.

Solution :

We have,

7×11×13+13

=13(7×11×1+1)

=13×78

=13×3×2×13

Hence, it is a composite number .

We have,               7×6×5×4×3×2×1+5

=5(7×6×4×3×2×1+1)

=5(1008+1)

=5×1009

Hence, it is a composite number .

Question 12:

There is a circular path around a sports field.Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time and go in the same direction, after how many minutes will they meet again at the starting point?

Solution :

Since, Both Sonia and Ravi move in the same direction and at the same time, the method to find the time when they will be meeting again at the starting point is LCM(18,12) is 2×3×3×2×1=36

Therefore, Sonia and Ravi will meet again after 36 minutes.

EXERCISE 1.3

Question 13:

Prove that 3–√ is irrational .

Solution :

Let us assume ,that 3–√ is rational

i.e 3–√=xy (where,x and y are co-primes)

y3–√=x

Squaring both sides

We get,(y3–√)2=x2

3y2=x2……..(1)

x2 is divisible by 3

So, x is also divisible by 3

 therefore we can write x=3k (for some values of k)

Substituting ,x=3k in eqn 1

3y2=(3k)2 y2=3k2

y2 is divisible by 3 it means y is divisible by 3

 therefore x and y are co-primes.

Since ,our assumption about 3–√ is rational is incorrect .

Hence, 3–√ is irrational number.

Question 14:

Prove that  3+23–√ is irrational .

Solution :

Let us assume that  3+23–√ is rational .

So, x= 3+23–√

x2=(3+23–√)2 x2=21+123–√

3–√=x2−2112…….(1)

 because x is a  rational number

So, the expression  x2−2112is also a rational number.This is a contradiction .Hence, 3+23–√ is irrational .

Question 15:

Prove that the following are irrational numbers.

i)13√

ii)73–√

iii)6+5–√

Solution :

i)13√

13√=3√3√3√

⇒ 13√=3√3

⇒ 13√=13×3–√

Let, a=13√=13×3–√ be a rational number .

⇒ 3a=3–√

3a is a rational number .Since product of any two rational numbers is a rational number which will imply that 3√ is a rational number .But,it contradicts since 3√ is a irrational number .

 therefore 3ais a irrational or ais irrational.

Hence, 13√ is irrational .

ii)73–√

Let, a=73–√ be a rational number .

⇒a7=3–√

a7 is a rational number .Since product of two rational number is a rational number. Which will imply that 7√

is a rational number .But,it contradicts since7√ is a irrational number .

 thereforea7is a irrational or ais irrational.

Hence,73–√ is irrational .

iii)6+5–√

Let, a=6+5–√ be a rational number .

Squaring , a2=(6+3–√)2

a2=36+23–√+3

a2=39+23–√

3–√=a2−3912……………..(1)

Since, ais a rational number

So, the expression  a2−3912 is also rational number.

⇒3–√ is a rational number .

This is a contradiction.

Hence, 6+5–√ is irrational.

EXERCISE 1.4

Question 16:

Without actually performing the long division, state whether the following rational numbers have a terminating decimal expansion or a non-terminating repeating decimal expansion.

i)233125

ii)2732

iii)3235

iv)51600

v)2949

vi)2724×53

vii)2922×56×73

viii)315

ix)3550

x)77210

Solution :

Note; If the denominator has only factors of 2 and 5 then it has terminating decimal expansion.

If the denominator has factors other than 2 and 5 then it has a non-terminating decimal expansion.

i)233125=2352

Since, the denominator has only 5 as its factor, it has a terminating decimal expansion.

ii)2732=2725

Since, the denominator has only 2 as its factor, it has a terminating decimal expansion.

iii)3235=323×7

Since, the denominator has factors other than 2 and 5, it has a non-terminating decimal expansion.

iv)51600=526×52

=126×51

Since the denominator has only  2 and 5 as its factors, it has a terminating decimal expansion.

v)2949=2972

Since, the denominator has factors other than 2 and 5, it has a non-terminating decimal expansion.

vi)2724×53

Since the denominator has only  2 and 5 as its factors, it has a terminating decimal expansion.

vii)2922×56×73

Since, the denominator has factors other than 2 and 5, it has a non-terminating decimal expansion.

viii)315=15

Since, the denominator has  only 5 as its factor, it has a terminating decimal expansion.

ix)3550=7×521×52

=721×51

Since the denominator has only  2 and 5 as its factors, it has a terminating decimal expansion.

x)77210=7×1121×51×3×7

=1121×51×3

Since, the denominator has factors other than 2 and 5, it has a non-terminating decimal expansion.

Question 17:

Write down the decimal expansion of the following rational numbers.

i)233125

ii)2732

iii)3235

iv)51600

v)2949

vi)2724×53

vii)2922×56×73

viii)315

ix)3550

x)77210

Solution;

i)233125=0.00736

ii)2732=.84375

iii)3235=.91428….

iv)51600=.00312

v)2949=.59183…

vi)2724×53=.0135

vii)2922×56×73=.00001…

viii)315=.2

ix)3550=.7

x)77210=.36¯

Question 18:

Decide whether the real numbers are rational or not. If they are rational ,then write its pq form.What can you say about the prime factors of q?

i)24.1352436789

ii).12346783940564543……

iii)41.2¯

Solution;

i)24.1352436789

Since it has a terminating decimal expansion, it is a rational number and q has factors of 2 and 5 only.

ii).12346783940564543……

Since, it has non-terminating and non- repeating decimal expansion, it is an irrational number.

iii)41.2¯

Since, it has non-terminating but repeating decimal expansion, it is a rational number and q has factors other than 2 and 5 .

SUMMARY ;

REAL NUMBER-Numbers that have a finite or infinite sequence of digits when it is represented in decimal form.

It is denoted by “R”

REAL NUMBERS ARE DIVIDED INTO 2 TYPES

RATIONAL NUMBERS- Number which can be represented in pq form.

It is divided into 2 types;

• TERMINATING DECIMAL NUMBER: The terminating decimal number has a finite number of Eg;0.432,653.8523,etc
• NON-TERMINATING RECURRING DECIMAL NUMBER: In a decimal, if a digit or a sequence of digits keeps repeating itself infinitely, then it is known as a non-terminating repeating decimal or recurring decimals.

Note;

It is expressed by putting a bar over the repeating digits.

eg;454.32¯,0.2¯,etc

IRRATIONAL NUMBERS- Number which can’t  be represented in pq form.

• NON- RECURRING DECIMAL NUMBER: It has infinite numbers and has the property that no sequence of digits are repeated.

eg-34.0428356393….,0.68384937…, etc.

 HENCE, We can conclude that for every real number there corresponds a unique point on the number line and conversely, to every point on the number line there corresponds a real number.

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