BlogCBSEExtra Questions for Class 9 Maths with Solutions Chapter Wise

Extra Questions for Class 9 Maths with Solutions Chapter Wise

Extra Questions for Class 9 Maths will help you with practice. We have selectively graded these extra questions for more practice. We request Students to solve these questions without going through solutions. If you face any difficulty in solving these Extra Questions, Please refer solutions.

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    Extra Questions for Class 9 Maths with Solutions

    Here is the list of Extra Questions for Class 9 Maths with Answers based on latest NCERT syllabus prescribed by CBSE.

    Number Systems Class 9 Extra Questions Very Short Answer Type

    Question 1.
    Simplify: (√5 + √2)2.
    Solution:
    Here, (√5 + √22 = (√52 + 2√5√2 + (√2)2
    = 5 + 2√10 + 2 = 7 + 2√10

    Question 2.
    Find the value of √(3)-2.
    Solution:
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 1

    Question 3.
    Identify a rational number among the following numbers :
    2 + √2, 2√2, 0 and π
    Solution:
    O is a rational number.

    Question 4.
    Express 1.8181… in the form pq where p and q are integers and q ≠ 0.
    Solution:
    Let x =1.8181… …(i)
    100x = 181.8181… …(ii) [multiplying eqn. (i) by 100] 99x = 180 [subtracting (i) from (ii)] x = 18099
    Hence, 1.8181… = 18099 = 2011

    Question 5.
    Simplify : √45 – 3√20 + 4√5
    Solution:
    √45 – 3√20 + 4√5 = 3√5 – 6√5 + 4√5 = √5.

    Question 6.
    Find the value of’
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 2
    Solution:
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 3

    Question 7.
    Find the value of
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 4
    Solution:
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 5

    Number Systems Class 9 Extra Questions Short Answer Type 1

    Question 1.
    Evaluate : (√5 + √22 + (√8 – √5)2
    Solution:
    (√5 + √2)2 + (√8 – √52 = 5 + 2 + 2√10 + 8 + 5 – 2√40
    = 20 + 2√10 – 4√10 = 20 – 2√10

    Question 2.
    Express 23.43¯¯¯¯¯ in pq form, where p, q are integers and q ≠ 0.
    Solution:
    Let x = 23.43¯¯¯¯¯
    or x = 23.4343… ….(i)
    100x = 2343.4343… …(ii) [Multiplying eqn. (i) by 100] 99x = 2320 [Subtracting (i) from (ii)
    ⇒ x = 232099
    Hence, 23.43¯¯¯¯¯ = 232099

    Question 3.
    Let ‘a’ be a non-zero rational number and ‘b’ be an irrational number. Is ‘ab’ necessarily an irrational ? Justify your answer with example.
    Solution:
    Yes, ‘ab’ is necessarily an irrational.
    For example, let a = 2 (a rational number) and b = √2 (an irrational number)
    If possible let ab = 2√2 is a rational number.
    Now, aba = 222 = √2 is a rational number.
    [∵ The quotient of two non-zero rational number is a rational] But this contradicts the fact that √2 is an irrational number.
    Thus, our supposition is wrong.
    Hence, ab is an irrational number.

    Question 4.
    Let x and y be a rational and irrational numbers. Is x + y necessarily an irrational number? Give an example in support of your answer.
    Solution:
    Yes, x + y is necessarily an irrational number.
    For example, let x = 3 (a rational number) and y = √5 (an irrational number)
    If possible let x + y = 3 + √5 be a rational number.
    Consider pq = 3 + √5, where p, q ∈ Z and q ≠ 0.
    Squaring both sides, we have
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 6
    pq is a rational
    ⇒ √5 is a rational
    But this contradicts the fact that √5 is an irrational number.
    Thus, our supposition is wrong.
    Hence, x + y is an irrational number.

    Number Systems Class 9 Extra Questions Short Answer Type 2

    Question 1.
    Represent √3 on the number line.
    Solution:
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 7
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 8
    On the number line, take OA = 1 unit. Draw AB = 1 unit perpendicular to OA. Join OB.
    Again, on OB, draw BC = 1 unit perpendicular to OB. Join OC.
    By Pythagoras Theorem, we obtain OC = √3. Using
    compasses, with centre O and radius OC, draw an arc, which intersects the number line at point
    D. Thus, OD = √3 and D corresponds to √3.

    Question 2.
    Represent √3.2 on the number line.
    Solution:
    First of all draw a line of length 3.2 units such that AB = 3.2 units. Now, from point B, mark a distance of 1 unit. Let this point be ‘C’. Let ‘O’ be the mid-point of the distance AC. Now, draw a semicircle with centre ‘O’ and radius OC. Let us draw a line perpendicular to AC passing through the point ‘B’ and intersecting the semicircle at point ‘D’.
    ∴ The distance BD = √3.2
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 9
    Now, to represent √3.2 on the number line. Let us take the line BC as number line and point ‘B’ as zero, point ‘C’ as ‘1’ and so on. Draw an arc with centre B and radius BD, which intersects the number line at point ‘E’.
    Then, the point ‘E’ represents √3.2.

    Question 3.
    Express 1.32 + 0.35 as a fraction in the simplest form.
    Solution:
    Let . x = 1.32 = 1.3222…..(i)

    Multiplying eq. (i) by 10, we have
    10x = 13.222…
    Again, multiplying eq. (i) by 100, we have
    100x = 132.222… …(iii)
    Subtracting eq. (ii) from (iii), we have
    100x – 10x = (132.222…) – (13.222…)
    90x = 119
    ⇒ x = 11990
    Again, y = 0.35 = 0.353535……
    Multiply (iv) by 100, we have …(iv)
    100y = 35.353535… (v)
    Subtracting (iv) from (u), we have
    100y – y = (35.353535…) – (0.353535…)
    99y = 35
    y = 3599
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 10

    Question 4.
    Find the square root of 10 + √24 + √60 + √40.
    Solution:
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 11

    Question 5.
    If x = 9 + 4√5, find the value of √x – 1x.
    Solution:
    Here,
    x = 9 + 4√5
    x = 5 + 4 + 2 x 2√5
    x = (√52 + (22 + 2 x 2x √5).
    x = (√5 + 2)2
    √x = √5 + 2
    Now, 1x = 15+2
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 12

    Question 6.
    If x = 152 , find the value of x3 – 32 – 5x + 3
    Solution:
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 13
    ∴ x – 2 = √5
    Squaring both sides, we have
    x2 – 4x + 4 = 5
    x2 – 4x – 1 = 0 …(i)
    Now, x3 – 32 – 5x + 3 = (x2 – 4x – 1) (x + 1) + 4
    = 0 (x + 1) + 4 = 4 [using (i)]

    Question 7.
    Find ‘x’, if 2x-7 × 5x-4 = 1250.
    Solution:
    We have 2x-7 × 5x-4 = 1250
    ⇒ 2x-7 × 5x-4 = 2 5 × 5 × 5 × 5
    ⇒ 2x-7 × 5x-4 = 21 × 54
    Equating the powers of 2 and 5 from both sides, we have
    ⇒ x – 7 = 1 and x – 4 = 4
    ⇒ x = 8 and x = 8
    Hence, x = 8 is the required value.

    Question 8.
    Evaluate:
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 14
    Solution:
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 15

    Number Systems Class 9 Extra Questions Long Answer Type

    `Question 1.
    If x = p+q+pqp+qpq, then prove that q2 – 2px + 9 = 0.
    Solution:
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 16
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 17
    Squaring both sides, we have
    ⇒ q2x2 + p2 – 2pqx = p2 – q2
    ⇒ q2x2 – 2pqx + q2 = 0
    ⇒ q(q2 – 2px + q) = 0
    ⇒ qx2 – 2px + q = 0 (∵ q ≠ 0)

    Question 2.
    If a = 1311 and b = 1a, then find a2 – b2
    Solution:
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 18

    Question 3.
    Simplify:
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 1.1
    Solution:
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 1.2

    Question 4.
    Prove that:
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 21
    Solution:
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 22

    Question 5.
    Find a and b, if
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 23
    Solution:
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 24

    Number Systems Class 9 Extra Questions HOTS

    Question 1.
    If xa = y, yb = z and zc = x, then prove that abc = 1.
    Solution:
    We have xa = y, yb = z and zc = x
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 25

    Question 2.
    Prove that:
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 26
    Solution:
    Taking L.H.S., we have
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 27

    Question 3.
    Show that:
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 28
    Solution:
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 29

    Number Systems Class 9 Extra Questions Value Based (VBQs)

    Question 1.
    Sudhir and Ashok participated in a long jump competition along a straight line marked as a number line. Both start the jumps one by one but in opposite directions. From ‘O’ Ashok jumps one unit towards the positive side while Sudhir jumps double in units as Ashok jumps, along negative side. After jumping 4 jumps each, at which point Ashok and Sudhir reached. What is the distance between their final positions ? Ashok argue that he is the winner since Sudhir is at negative side. Who do you think is winner and why? What is the value of the competition ?
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 30
    Solution:
    After jumping four jumps each, Ashok reached at 4 in positive direction and Sudhir reached at -8 i.e., in negative direction. Distance between their final positions is 12 units. Here, distance covered by Sudhir is 8 units and distance covered by Ashok is 4 units. Thus, Sudhir is the winner. Competition inculcate spirit of performance.

    Question 2.
    Manu went to his mathematics teacher and asked him “Sir, I want some chocolates to distribute among my classmates for my birthday but I have no money. Can you provide me some chocolates”. Teacher told Manu, I am giving you two numbers 13+22 and 1322 and if you can find the value of sum of their squares, then I will provide you as many chocolates as the resulting value of sum of squares of given numbers. Find the number of chocolates. What value is depicted from this action?
    Solution:
    Number Systems Class 9 Extra Questions Maths Chapter 1 with Solutions Answers 31
    = (3 – 2√2)2 + (3 + 2√22
    = 9 + 8 – 2 × 3 × (2√2) + 9 + 8 + 2 × 3 × 2√2 = 34.
    Hence, resulting value of sum of squares of numbers = number of chocolates = 34. By doing this, teacher motivates the students to use their knowledge and apply it in day to day life with caring and kindness.

    Polynomials Class 9 Extra Questions Very Short Answer Type

    Question 1.
    Factorise : 125x3 – 64y3
    Solution:
    125x3– 6443 = (5x)3 – (4y)3
    By using a3 – b3 = (a – b) (a2 + ab + b2), we obtain
    125x3– 64y3 = (5x – 4y) (25x2 + 20xy + 16y2)

    Question 2.
    Find the value of (x + y)2 + (x – y)2.
    Solution:
    (x + y)2 + (x – y)2 = x2 + y2 + 2xy + x2 + y2 – 2xy
    = 2x2 + 2y2 = 21x2 + y2)

    It is a free online tool that shows the division of two polynomials with Polynomial long division method.

    Question 3.
    If p(x)= x2 – 2√2x+1, then find the value of p(2√2)
    Solution:
    Put x = 2√2 in p(x), we obtain
    p(2√2) = (2√2)2 – 2√2(2√2) + 1 = (2√2)2 – (2√2)2 + 1 = 1

    Question 4.
    Find the value of m, if x + 4 is a factor of the polynomial x2 + 3x + m.
    Solution:
    Let p(x) = x2 + 3x + m
    Since (x + 4) or (x – (-4)} is a factor of p(x).
    ∴ p(-4) = 0
    ⇒ (-4)2 + 3(-4) + m = 0
    ⇒ 16 – 12 + m = 0
    ⇒ m = -4

    Question 5.
    Find the remainder when x3+ x2 + x + 1 is divided by x – 12 using remainder theorem.
    Solution:
    Let p(x) = x3+ x2 + x + 1 and q(x) = x – 12
    Here, p(x) is divided by q(x)
    ∴ By using remainder theorem, we have
    Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 1

    Question 6.
    Find the common factor in the quadratic polynomials x2 + 8x + 15 and x2 + 3x – 10.
    Solution:
    x2 + 8x + 15 = x2 + 5x + 3x + 15 = (x + 3) (x + 5)
    x2 + 3x – 10 = x2 + 5x – 2x – 10 = (x – 2) (x + 5)
    Clearly, the common factor is x + 5.

    Polynomials Class 9 Extra Questions Short Answer Type 1

    Question 1.
    Expand :
    (i) (y – √3)2
    (ii) (x – 2y – 3z)2
    Solution: (i)
    (y – √3)2 = y2 -2 × y × √3 + (√3)2 = y2 – 2√3 y + 3 (x – 2y – 3z)2
    = x2 + 1 – 2y)2 + (-3z)2 + 2 × x × (-2y) + 2 × (-2y) × (-3z) + 2 × (-3z) × x
    = x2 + 4y2 + 9z2 – 4xy + 12yz – 6zx

    Question 2.
    If x + = 1x = 7, then find the value of x3 + 1x3
    Solution:
    We have x + 1x = 7
    Cubing both sides, we have
    Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 2

    Question 3.
    Show that p – 1 is a factor of p10 + p8 + p6 – p4 – p2 – 1.
    Solution:
    Let f(p) = p10 + p8 + p6 – p4 – p2 – 1
    Put p = 1, we obtain
    f(1) = 110 + 18 + 16 – 14 – 12 – 1
    = 1 + 1 + 1 – 1 – 1 – 1 = 0
    Hence, p – 1 is a factor of p10 + p8 + p6 – p4 – p2 – 1.

    Question 4.
    If 3x + 2y = 12 and xy = 6, find the value of 27x3 + 8y3
    Solution:
    We have 3x + 2y = 12
    On cubing both sides, we have
    ⇒ (3x + 2y)3 = 123
    ⇒ (3x)3 +(2y)3 + 3 × 3x × 2y(3x + 2y) = √728
    ⇒ 27x3+ 8y3 + 18xy(3x + 2y) = √728
    ⇒ 27x3+ 8y3 + 18 × 6 × 12 = √728
    ⇒ 27x3+ 8y3 + 1296 = √728
    ⇒ 27x3+ 8y3 = √728 – 1296
    ⇒ 27x3+ 8y3 = 432

    Question 5.
    Factorise : 4x2 + 9y2 + 16z22 + 12xy – 24 yz – 16xz.
    Solution:
    4x2 + 9y2 + 16z22 + 12xy – 24yz – 16xz
    = (2x)2 + (3y)2 + (-4z)2 + 2(2x)(3y) + 2(3y)(= 42) + 2(- 42)(2x)
    By using a2 + b2 + 2ab + 2bc + 2ca = (a + b + c)2, we obtain
    = (2x + 3y – 4z)2 = (2x + 3y – 4z) (2x + 3y – 4z)

    Question 6.
    Factorise : 1 – 2ab – (a2 + b2).
    Solution:
    1 – 2ab – (a2 + b2) = 1 – (a2 + b2 + 2ab)
    = 12 – (a + b)2
    = (1 + a + b) (1 – a – b) [∵ x2 – y2 = (x + y)(x – y)]

    Polynomials Class 9 Extra Questions Short Answer Type 2

    Question 1.
    Factorise :
    Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 3
    Solution:
    Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 4

    Question 2.
    Factorise 64a3 – 27b3 – 144a2b + 108ab2.
    Solution:
    64a2 – 27b2 – 144a2b + 108ab2
    = (4a)3 – (3b)3 – 36ab(4a – 3b)
    = (40)2 – (3b)3 – 3 × 4a × 3b (4a – 3b)
    = (4a – 3b)3 [∵ (x – y)3 = x3 – y3 – 3xy(x – y)] = (40 – 3b) (4a – 3b) (4a – 3b)

    Question 3.
    What are the possible expressions for the dimensions of a cuboid whose volume is given below ?
    Volume = 12ky2 + 8ky – 20k.
    Solution:
    We have, volume = 12ky2 + 8ky – 20k
    = 4k(3y2 + 2y – 5) = 4k(3y2 + 5y – 3y – 5)
    = 4k[y(3y + 5) – 1(3y + 5)] = 4k(3y + 5) (y – 1)
    ∴Possible expressions for the dimensions of cuboid are 4k units, (3y + 5) units and (y – 1) units.

    Question 4.
    If p(x) = x3 + 3x2 – 2x + 4, then find the value of p(2) + p(-2) – P(0).
    Solution:
    Here, p(x) = x3+ 3x2 – 2x + 4
    Now, p(2) = 23 + 3(2)2 – 2(2) + 4
    = 8 + 12 – 4 + 4 = 20
    p(-2) = (-2)3 + 3(-2)2 – 21 – 2) + 4
    = 8 + 12 + 4 + 4 = 12
    and p(0) = 0 + 0 – 0 + 4 = 4
    ∴ p(2) + p(-2) – p(0) = 20 + 12 – 4 = 28.

    Question 5.
    If one zero of the polynomial x2 – √3x + 40 is 5, which is the other zero ?
    Solution:
    Let p(x) = x2 – √3x + 40
    = x2 – 5x – 8x + 40 = x(x – 5) – 8(x – 5) = (x – 5) (x – 8)
    Now, for zeroes of given polynomial, put p(x) = 0
    ∴ (x – 5) (x – 8) = 0
    ⇒ x = 5 or x = 8
    Hence, other zero is 8.

    Question 6.
    Simplify:
    Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 5
    Solution:
    Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 6

    Question 7.
    If one zero of the polynomial x2 – √3x + 40 is 5, which is the other zero ?
    Solution:
    Let
    p(x) = x2 – √3x + 40
    = x2 – 5x – 8x + 40 = x(x – 5) – 8(x – 5) = (x – 5) (x – 8)
    Now, for zeroes of given polynomial, put p(x) = 0
    ∴ (x – 5) (x – 8) = 0
    x = 5 or x = 8
    ⇒ Hence, other zero is 8.

    Polynomials Class 9 Extra Questions Long Answer Type

    Question 1.
    Prove that (a + b + c)3 – a3 – b3 – c3 = 3(a + b) (b + c) (c + a).
    Solution:
    L.H.S. = (a + b + c)3 – a3 – b3 – c3
    = {(a + b + c)3 – 3} – {b3 + c3}
    = (a + b + c – a) {(a + b + c)2 + a2 + a(a + b + c)} – (b + c) (b2 + c2 – bc)
    = (b + c) {a2 + b2 + 2 + 2ab + 2bc + 2ca + a2 + a2 + ab + ac – b2 – a2 + bc)
    = (b + c) (3a2 + 3ab + 3bc + 3ca}
    = 3(b + c) {a2 + ab + bc + ca}
    = 31b + c) {{a2 + ca) + (ab + bc)}
    = 3(b + c) {a(a + c) + b(a + c)}
    = 3(b + c)(a + c) (a + b)
    = 3(a + b)(b + c) (c + a) = R.H.S.

    Question 2.
    Factorise : (m + 2n)2 x2 – 22x (m + 2n) + 72.
    Solution:
    Let m + 2n = a
    ∴ (m + 2n)2 x2 – 22x (m + 2n) + 72 = a2x2 – 22ax + 72
    = a2x2 – 18ax – 4ax + 72
    = ax(ax – 18) – 4(ax – 18)
    = (ax – 4) (ax – 18)
    = {(m + 2n)x – 4)} {(m + 2n)x – 18)}
    = (mx + 2nx – 4) (mx + 2nx – 18).

    Question 3.
    If x – 3 is a factor of x2 – 6x + 12, then find the value of k. Also, find the other factor of the – polynomial for this value of k.
    Solution:
    Here, x – 3 is a factor of x2 – kx + 12
    ∴ By factor theorem, putting x = 3, we have remainder 0.
    ⇒ (3)2 – k(3) + 12 = 0
    ⇒ 9 – 3k + 12 = 0
    ⇒ 3k = 21
    ⇒ k = 7
    Now, x2 – 7x + 12 = x2 – 3x – 4x + 12
    = x(x – 3) – 4(x – 3)
    = (x – 3) (x – 4)
    Hence, the value of k is 7 and other factor is x – 4.

    Question 4.
    Find a and b so that the polynomial x3– 10x2 + ax + b is exactly divisible by the polynomials (x – 1) and (x – 2).
    Solution:
    Let p(x) = x3– 10x2 + ax + b
    Since p(x) is exactly divisible by the polynomials (x – 1) and (x – 2).
    ∴ By putting x = 1, we obtain
    (1)3 – 10(1)2 + a(1) + b = 0
    ⇒ a + b = 9
    And by putting x = 2, we obtain
    (2)3 – 10(2)2 + a(2) + b = 0
    8 – 40 + 2a + b = 0
    ⇒ 2a + b = 32
    Subtracting (i) from (ii), we have
    a = 23
    From (i), we have 23 + b = 9 = b = -14
    Hence, the values of a and b are a = 23 and b = -14

    Question 5.
    Factorise : x2 – 6x2 + 11x – 6.
    Solution:
    Let p(x) = x2 – 6x2 + 11x – 6
    Here, constant term of p(x) is -6 and factors of -6 are ± 1, ± 2, ± 3 and ± 6
    By putting x = 1, we have
    p(1) = (1)3 – 6(1)2 + 11(1) – 6 = 1 – 6 + 11 -6 = 0
    ∴ (x – 1) is a factor of p(x)
    By putting x = 2, we have
    p(2) = (2)3 – 6(2)2 + 11(2) – 6 = 8 – 24 + 22 – 6 = 0
    ∴ (x – 2) is a factor of p(x)
    By putting x = 3, we have
    p(3) = (3)3 – 6(3)2 + 11(3) – 6 = 27 – 54 + 33 – 6 = 0
    ∴ (x – 3) is a factor of p(x) Since p(x) is a polynomial of degree 3, so it cannot have more than three linear factors.
    ∴ x3 – 6x2 + 11x – 6 = k (x – 1) (x – 2) (x – 3)
    By putting x = 0, we obtain
    0 – 0 + 0 – 6 = k (-1) (-2) (3)
    -6 = -6k
    k = 1
    Hence, x3 – 6x2 + 11x – 6 = (x – 1) (x – 2)(x – 3).

    Question 6.
    Show that 13and 43 are zeroes of the polynomial 9x3 – 6x2 – 11x + 4. Also, find the third zero of the polynomial.
    Solution:
    Let p(x) = 9x3– 6x2 – 11x + 4
    Put x = 13, we have
    Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 7
    Thus, x = 43 is another zero of the polynomial p(x). Since x = 13 and x = 43 are the zeroes of p(x), therefore, (x13) (x43) (3x – 1) (3x – 4) or 9x2 – 15x + 4 exactly divides p(x).
    ⇒ 9x3 – 6x2 – 11x + 4 = (9x2 – 15x + 4) (x + 1)
    Hence, x = -1 is its third zero.

    Question 7.
    Factorise : 6x2 – 5x2 – √3x + 12
    Solution:
    Let p(x) = 6x3– 5x2 – √3x + 12
    Here, constant term of p(x) is 12 and factors of 12 are ± 1, ± 2, ± 3, ± 4, ± 6, ± 12.
    By putting x = 1, we have
    p(1) = 6(1)3 – 5(1)2 – √3(1) + 12 = 6 – 5 – √3 + 12 = 0
    ∴ (x – 1) is a factor of p(x).
    Now, by long division, we have
    Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 8
    Thus,
    p(x) = (x – 1) (6x2 + x – 12)
    p(x) = (x – 1) (6x2 + 9x – 8x – 12)
    p(x) = (x – 1) {3x (2x + 3) – 4(2x + 3)}
    p(x) = (x – 1) (3x – 4) (2x + 3).

    Also check: NCERT Solutions Class 9 Maths Chapter 12 – Heron’s Formula

    Polynomials Class 9 Extra Questions HOTS

    Question 1.
    What must be added to polynomial f(x) = x4 + 2x2 – 2x2 + x – 1 so that resulting polynomial is exactly divisible by x2 + 2x – 3?
    Solution:
    Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 9
    Here, remainder = -x + 2
    To make remainder = 0, we must add -(remainder) in the polynomial
    i.e., -(-x + 2) i.e., x – 2
    Hence, x4 + 2x3 – 2x2 + x – 1 + (x – 2)
    Here, polynomial = x4 + 2x3– 2x2 + 2x – 3 and required addition is (x – 2).

    Question 2.
    If x = 2 – √3, y = √3 – √7 and 2 = √7 – √4, find the value of x’ + 43 + 2?.
    Solution:
    Here, x + y + z = 2 – √3+ √3 – √7+√7 – 2 = 0
    x3+ √3 + x3= 3(x)(y)(z)
    = 3(2 – √3)(√3 – √7)(√7 – 2)
    = 3(2√3 – 2√7 – 3 + √21)(√7 – 2)
    = 3(2√21 – 14 – 3√7 + 7√3 – 4√3 + 4√7 + 6 – 2√21)
    = 3(3√3 + √7 – 8)

    Question 3.
    If (x – a) is a factor of the polynomials x2 + px – q and x2 + rx – t, then prove that a = tqrp
    Solution:
    Let f(x) = x + px -q and g(x) = x2 + x – t
    Since x-a is factor of both f(x) and g(x)
    ⇒ f(a) = g(a) = 0
    Now, here f(a) = a2 + pa – q and
    g(a) = a2 + ra- t
    ⇒ a2 + pa – q = a + ra – t (considering f(a) = g(a)] ⇒ pa – q = ra – t
    ⇒ ra – pa = t – q
    ⇒ a(r – p) = t – q
    a = tqrp

    Polynomials Class 9 Extra Questions Value Based (VBQs)

    Question 1.
    If a teacher divides a material of volume 27x3 + 54x2 + 36x + 8 cubic units among three students. Is it possible to find the quantity of material ? Can you name the shape of the figure teacher obtained ? Which value is depicted by the teacher ?
    Solution:
    We know that, √olume = Length × Breadth × Height
    Now, 27x3+ 54x2 + 36x + 8
    = (3x)3 + 3(3x)2(2) + 3(3x)(2)2 + (2)3
    = (3x + 2)2 = (3x + 2) (3x + 2) (3x + 2)
    Thus, volume = (3x + 2) (3x + 2) (3x + 2)
    Yes, it is possible to find the quantity of material. (3x + 2) units.
    Cube.
    Apply knowledge and use of example for clarity of subject, student friendly.

    Question 2.
    In a camp organised by the students of class-9 to donate amount collected to flood victims of Kashmir. At the time of payment of a juice glass at one stall of juice, stall holder asked the students to pay the remainder of x3+ 3x2 + 3x + 1 divided by (x12) What is the price of the juice at the stall ? Which value is depicted by class-9 students by organising such camps ?
    Solution:
    Let
    p(x) = x2 + 3x2 + 3x + 1 and
    Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 10
    By long division method, we have
    Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 11
    Remainder = 278 or 3 38
    Thus, price of the juice glass is ₹ 3 38
    Caring, kindness, social welfare and helping in development of the needy.

    Related Links:

    Polynomials Class 9 Extra Questions Maths Chapter 2

    NCERT Books for Class 9 – Free PDF Download for 2023-24

    NCERT Solutions Class 9 Science

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