Polynomials -Exercise 2.2 – Class 10

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. 1/4 , -1
  2. √2, 1/3
  3. 0, √5
  4. 1,1
  5. 1/4 , ¼
  6. 4, 1

Solution:

  1. 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


 

 

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