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

---