**1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. **

**(i) x ^{2} – 2x – 8**

**(ii) 4s ^{2} – 4s + 1**

**(iii) 6x ^{2} – 3 – 7x**

**(iv) 4u ^{2} + 8u**

**(v) t ^{2} – 15**

**(vi) 3x ^{2} – x – 4**

Solution:

(i) x^{2} – 2x – 8 = (x – 4)(x+2)

The value of x^{2} – 2x – 8 is zero when x − 4 = 0 or x + 2 = 0, i.e., when x = 4 or x = −2

Therefore, the zeros of x^{2} – 2x – 8 are 4 and -2.

Sum of zeros = 4 – 2 = 2 = –^{(-2)}/_{1} = – (coefficient of x)/(coefficient of x^{2})

Product of zeros = 4 x (-2) = -8 = (-8)/1 = (constant term/coefficient of x^{2}

(ii) 4s^{2} – 4s + 1 = (2s – 1)^{2}

The value of 4s^{2} – 4s + 1 is zero when 2s – 1 = 0 or s = 1/2,

Therefore, the zeros of 4s^{2} – 4s + 1 are ^{1}/_{2} and ^{1}/_{2}.

Sum of zeros = ^{1}/_{2 }+ ^{1}/_{2 }= 1 = –^{(-4)}/_{4} = – (coefficient of s)/(coefficient of s^{2})

Product of zeros = ^{1}/_{2 }x ^{1}/_{2 } = ^{1}/_{4 } = (constant term/coefficient of s^{2}

^{ }

(iii) 6x^{2} – 3 – 7x = 6x^{2} – 7x – 3 = (3x + 1)(2x – 3)

The value of 6x^{2} – 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 6x^{2} – 3 – 7x are –^{1}/_{3} and ^{3}/_{2}.

Sum of zeros = –^{1}/_{3} + ^{3}/_{2 }= ^{7}/_{6} = –^{(-7)}/_{6} = – (coefficient of x)/(coefficient of x^{2})

Product of zeros = –^{1}/_{3} x ^{3}/_{2} = – ^{1}/_{2} = ^{-3}/_{6} = (constant term/coefficient of x^{2}

(iv) 4u^{2} + 8u = 4u^{2} + 8u + 0 = 4u(u + 2)

The value of 4u^{2} + 8u is zero when 4u = 0 or u + 2 = 0, i.e., when u = 0 or u = −2

Therefore, the zeros of 4u^{2} + 8u are 0 and -2.

Sum of zeros = 0 + (-2) = -2 = –^{(-8)}/_{4} = – (coefficient of u)/(coefficient of u^{2})

Product of zeros = 0 x (-2) = 0 = ^{0}/_{4 }= (constant term/coefficient of u^{2}

(v) t^{2} – 15 = t^{2} – 0.t – 15 = (t – √15)(t + √15)

The value of t^{2} – 15 is zero when t – √(15) = 0 or t + √(15) = 0, i.e., when t = √15 or t = −√15

Therefore, the zeros of t^{2} – 15 are √15 and -√15

Sum of zeros = √15 + (-√15) = 0 = –^{(-0)}/_{1} = – (coefficient of t)/(coefficient of t^{2})

Product of zeros = √15 x (-√15) = -15 = ^{-√15}/_{1} = (constant term/coefficient of t^{2}

(vi) 3x^{2} – x – 4 = (3x – 4)(x + 1)

The value of 3x^{2} – 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 3x^{2} – x – 4 are ^{4}/_{3} and -1.

Sum of zeros = ^{4}/_{3} + (-1) = ^{1}/_{3} = –^{(-1)}/_{3} = – (coefficient of x)/(coefficient of x^{2})

Product of zeros = ^{4}/_{3} (-1) = –^{4}/_{3} = (constant term)/(coefficient of x^{2)}

**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 ax^{2} + 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 4x^{2} – x – 4

(2) √2, ^{1}/_{3}

Let the polynomial be ax^{2} + 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 x^{2} + √5

_{(3) 0, }√5

Let the polynomial be ax^{2} + 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 3x^{2} – 3√2 x + 1

(4) 1,1

Let the polynomial be ax^{2} + 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 x^{2} – x + 1

(5) –^{1}/_{4 }, ^{¼}

Let the polynomial be ax^{2} + 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 4x^{2} + x + 1

(6) 4, 1

Let the polynomial be ax^{2} + 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 x^{2} – 4x + 1

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