1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x2 – 2x – 8
(ii) 4s2 – 4s + 1
(iii) 6x2 – 3 – 7x
(iv) 4u2 + 8u
(v) t2 – 15
(vi) 3x2 – x – 4
Solution:
(i) x2 – 2x – 8 = (x – 4)(x+2)
The value of x2 – 2x – 8 is zero when x − 4 = 0 or x + 2 = 0, i.e., when x = 4 or x = −2
Therefore, the zeros of x2 – 2x – 8 are 4 and -2.
Sum of zeros = 4 – 2 = 2 = –(-2)/1 = – (coefficient of x)/(coefficient of x2)
Product of zeros = 4 x (-2) = -8 = (-8)/1 = (constant term/coefficient of x2
(ii) 4s2 – 4s + 1 = (2s – 1)2
The value of 4s2 – 4s + 1 is zero when 2s – 1 = 0 or s = 1/2,
Therefore, the zeros of 4s2 – 4s + 1 are 1/2 and 1/2.
Sum of zeros = 1/2 + 1/2 = 1 = –(-4)/4 = – (coefficient of s)/(coefficient of s2)
Product of zeros = 1/2 x 1/2 = 1/4 = (constant term/coefficient of s2
(iii) 6x2 – 3 – 7x = 6x2 – 7x – 3 = (3x + 1)(2x – 3)
The value of 6x2 – 3 – 7x is zero when 3x + 1 = 0 or 2x – 3 = 0, i.e., when x = –1/3 or x = 3/2
Therefore, the zeros of 6x2 – 3 – 7x are –1/3 and 3/2.
Sum of zeros = –1/3 + 3/2 = 7/6 = –(-7)/6 = – (coefficient of x)/(coefficient of x2)
Product of zeros = –1/3 x 3/2 = – 1/2 = -3/6 = (constant term/coefficient of x2
(iv) 4u2 + 8u = 4u2 + 8u + 0 = 4u(u + 2)
The value of 4u2 + 8u is zero when 4u = 0 or u + 2 = 0, i.e., when u = 0 or u = −2
Therefore, the zeros of 4u2 + 8u are 0 and -2.
Sum of zeros = 0 + (-2) = -2 = –(-8)/4 = – (coefficient of u)/(coefficient of u2)
Product of zeros = 0 x (-2) = 0 = 0/4 = (constant term/coefficient of u2
(v) t2 – 15 = t2 – 0.t – 15 = (t – √15)(t + √15)
The value of t2 – 15 is zero when t – √(15) = 0 or t + √(15) = 0, i.e., when t = √15 or t = −√15
Therefore, the zeros of t2 – 15 are √15 and -√15
Sum of zeros = √15 + (-√15) = 0 = –(-0)/1 = – (coefficient of t)/(coefficient of t2)
Product of zeros = √15 x (-√15) = -15 = -√15/1 = (constant term/coefficient of t2
(vi) 3x2 – x – 4 = (3x – 4)(x + 1)
The value of 3x2 – x – 4 is zero when 3x − 4 = 0 or x + 1 = 0, i.e., when x = 4/3 or x = −1
Therefore, the zeros of 3x2 – x – 4 are 4/3 and -1.
Sum of zeros = 4/3 + (-1) = 1/3 = –(-1)/3 = – (coefficient of x)/(coefficient of x2)
Product of zeros = 4/3 (-1) = –4/3 = (constant term)/(coefficient of x2)
2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively
- 1/4 , -1
- √2, 1/3
- 0, √5
- 1,1
- –1/4 , ¼
- 4, 1
Solution:
- 1/4 , -1
Let the polynomial be ax2 + bx + c and its zeroes be 𝛼 and ᵝ
𝛼 + ᵝ = 1/4 = –b/a
𝛼 x ᵝ = -1 = -4/4 = c/a
If a = 4, then b = -1, c = -4
Therefore, the quadratic polynomial is 4x2 – x – 4
(2) √2, 1/3
Let the polynomial be ax2 + bx + c and its zeroes be 𝛼 and ᵝ
𝛼 + ᵝ =0 = 0/1 = –b/a
𝛼 x ᵝ = √5 = – √5/1 = c/a
If a = 1, then b = 0, c = √5
Therefore, the quadratic polynomial is x2 + √5
(3) 0, √5
Let the polynomial be ax2 + bx + c and its zeroes be 𝛼 and ᵝ
𝛼 + ᵝ =0 = √2 =-3√2/3 = –b/a
𝛼 x ᵝ = 1/3 = c/a
If a = 3, then b = -3 √2, c =1
Therefore, the quadratic polynomial is 3x2 – 3√2 x + 1
(4) 1,1
Let the polynomial be ax2 + bx + c and its zeroes be 𝛼 and ᵝ
𝛼 + ᵝ =1 = 1/1 = –b/a
𝛼 x ᵝ = 1 = 1/1 = c/a
If a = 1, then b = -1, c = 1
Therefore, the quadratic polynomial is x2 – x + 1
(5) –1/4 , ¼
Let the polynomial be ax2 + bx + c and its zeroes be 𝛼 and ᵝ
𝛼 + ᵝ = -1/4 = –b/a
𝛼 x ᵝ = 1/4 = = c/a
If a = 4, then b = 1, c = 1
Therefore, the quadratic polynomial is 4x2 + x + 1
(6) 4, 1
Let the polynomial be ax2 + bx + c and its zeroes be 𝛼 and ᵝ
𝛼 + ᵝ = 4 = 4/1 = –b/a
𝛼 x ᵝ = 1 = 1/1 = c/a
If a = 1, then b = -4, c = 1
Therefore, the quadratic polynomial is x2 – 4x + 1
1 thought on “Polynomials -Exercise 2.2 – Class 10”
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