MATHEMATICS CLASS 10TH EXERCISE 2.2 SOLUTION ncert

MATHEMATICS CLASS 10TH EXERCISE 2.2 SOLUTION

1. Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively

  1. x2 – 2x – 8
    x2 – 2x – 8
    = x2 – 4x + 2x – 8
    = x(x – 4) + 2(x – 4)
    = (x + 2)(x – 4)
    Now, x + 2 = 0
    ⇒ x = -2;
    x – 4 = 0
    ⇒ x = 4
    So, zeroes are x = -2 and 4.
    Verification:
    α = -2, β = 4
    α + β = -2 + 4 = 2 = -coefficient of x / coefficient of x2 = -(-2)/1 = 2 (verified)
    αβ = (-2) × 4 = -8 = constant term/coefficient of x2 = -8/1 = -8 (verified)
  2. 4s2 – 4s + 1
    4s2 – 4s + 1
    = 4s2 – 2s – 2s + 1
    = 2s(2s – 1) – 1(2s – 1)
    = (2s – 1)(2s – 1)
    Now, 2s – 1 = 0
    ⇒ s = 1/2
    So, zeroes are s = 1/2 and 1/2.
    Verification:
    α = 1/2, β = 1/2
    α + β = 1/2 + 1/2 = 1 = -coefficient of s / coefficient of s2 = -(-4)/4 = 1 (verified)
    αβ = (1/2) × (1/2) = 1/4 = constant term/coefficient of s2 = 1/4 (verified)
  3. 6x2 – 7x – 3
    6x2 – 7x – 3
    = 6x2 – 9x + 2x – 3
    = 3x(2x – 3) + 1(2x – 3)
    = (3x + 1)(2x – 3)
    Now, 3x + 1 = 0
    ⇒ x = -1/3
    2x – 3 = 0
    ⇒ x = 3/2
    So, zeroes are x = -1/3 and 3/2.
    Verification:
    α = -1/3, β = 3/2
    α + β = -1/3 + 3/2 = 7/6 = -coefficient of x / coefficient of x2 = -(-7)/6 = 7/6 (verified)
    αβ = (-1/3) × (3/2) = -1/2 = constant term/coefficient of x2 = -3/6 = -1/2 (verified)
  4. 4u2 + 8u
    4u2 + 8u
    = 4u(u + 2)
    Now 4u = 0
    ⇒ u = 0
    u + 2 = 0
    ⇒ u = -2
    So, zeroes are u = 0 and -2.
    Verification:
    α = 0, β = -2
    α + β = 0 + (-2) = -2 = -coefficient of u / coefficient of u2 = -8/4 = -2 (verified)
    αβ = 0 × (-2) = 0 = constant term/coefficient of u2 = 0/4 = 0 (verified)
  5. t2 – 15
    t2 – 15
    = (t – √15)(t + √15)
    Now, t – √15 = 0
    ⇒ t = √15
    t + √15 = 0
    ⇒ t = -√15
    So, zeroes are t = √15 and -√15.
    Verification:
    α = √15, β = -√15
    α + β = √15 + (-√15) = 0 = -coefficient of t / coefficient of t2 = 0/1 = 0 (verified)
    αβ = (√15) × (-√15) = -15 = constant term/coefficient of t2 = -15/1 = -15 (verified)
  6. 3x2 – x – 4
    3x2 – x – 4
    = 3x2 – 3x + 4x – 4
    = 3x(x + 1) – 4(x + 1)
    = (3x – 4)(x + 1)
    Now, 3x – 4 = 0
    ⇒ x = 4/3
    x + 1 = 0
    ⇒ x = -1
    So, zeroes are x = 4/3 and -1.
    Verification:
    α = 4/3, β = -1
    α + β = 4/3 + (-1) = 1/3 = -coefficient of x / coefficient of x2 = -(-1)/3 = 1/3 (verified)
    αβ = (4/3) × (-1) = -4/3 = constant term/coefficient of x2 = -4/3 (verified)

2. Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively

  1. Sum = 14, Product = -1
    Let α and β be the zeroes.
    Polynomial: x2 – (α + β)x + αβ = x2 – 14x – 1
  2. Sum = √2, Product = 13
    Let α and β be the zeroes.
    Polynomial: x2 – (√2)x + 13
  3. Sum = 0, Product = √5
    Let α and β be the zeroes.
    Polynomial: x2 + √5
  4. Sum = 1, Product = 1
    Let α and β be the zeroes.
    Polynomial: x2 – x + 1
  5. Sum = -14, Product = 14
    Let α and β be the zeroes.
    Polynomial: x2 + 14x + 14
  6. Sum = 4, Product = 1
    Let α and β be the zeroes.
    Polynomial: x2 – 4x + 1

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