Exercise 2.2

(i)

(i) x2– 2x – 8

Apply the Quadratic Formula : a = , b = , c = .

Then, α and β are the solutions obtained by applying the (+) and (-) signs, respectively.

α =;β = .

Sum of zeroes =α + β=-ba.

Here, α + β = - 2 + 4 = ; - ba = - (- 2) / 1 = .

Hence, sum of the zeroes (α + β) = - ba is verified.

Product of Zeroes= α × β = ca.

α × β = - 2 × 4 = ; ca = - 8 / 1 = .

Hence, product of zeroes (α × β) = ca is verified.

(ii)

(ii) 4s2 – 4s + 1

Apply the Quadratic Formula : a = , b = , c = .

Then, α and β are the solutions obtained by applying the (+) and (-) signs, respectively.

α =;β = .

Sum of Zeroes = α + β=−ba.

Here, α + β = 12 + 12 = ; - ba = - −44 = .

Hence, sum of the zeroes (α + β) = - ba is verified.

Product of Zeroes = α × β = ca.

α × β = 12 × 12 = 1/4-1/4; ca = 14 = 1/4-1/4.

Hence, product of zeroes (α × β) = ca is verified.

(iii)

(iii) 6x2−3−7x

Apply the Quadratic Formula : a = , b = , c = .

Then, α and β are the solutions obtained by applying the (+) and (-) signs, respectively.

α =3223;β = −133.

Sum of Zeroes = α + β=−ba.

Here, α + β = 32 + −13 = ; - ba = - −76 = .

Hence, sum of the zeroes (α + β) = - ba is verified.

Product of Zeroes = α × β = ca.

α × β = 32 × −13 = = -1/21/2; ca = −36 = -1/21/2.

Hence, product of zeroes (α × β) = ca is verified.

(iv)

(iv) 4u2+ 8u

Apply the Quadratic Formula : a = , b = , c = .

Then, α and β are the solutions obtained by applying the (+) and (-) signs, respectively.

α =01;β = -2-3.

Sum of Zeroes = α + β=−ba.

Here, α + β = 0 + (- 2) = ; -ba = - 84 = -22.

Hence, sum of the zeroes (α + β) = - ba is verified.

Product of Zeroes = α × β = ca.

α × β = 0 × (-2) = = ; ca = 04 = .

Hence, product of zeroes (α × β) = ca is verified.

(v)

(v) t2– 15

Apply the Quadratic Formula : a = , b = , c = .

Then, α and β are the solutions obtained by applying the (+) and (-) signs, respectively.

α =15−15;β = −15−−15.

Sum of Zeroes = α + β=−ba.

Here, α + β = 15 + (−15) = ; -ba = - 01 = .

Hence, sum of the zeroes (α + β) = - ba is verified.

Product of Zeroes = α × β = ca.

α × β = 15 × (−15) = = ; ca = −151 = .

Hence, product of zeroes (α × β) = ca is verified.

(vi)

(vi) 3x2 – x – 4

Apply the Quadratic Formula : a = , b = , c = .

Then, α and β are the solutions obtained by applying the (+) and (-) signs, respectively.

α =4/3-4/3;β = .

Sum of Zeroes = α + β=−ba.

Here, α + β = 43 + (-1) = ; -ba = - −13 = .

Hence, sum of the zeroes (α + β) = - ba is verified.

Product of Zeroes = α × β = ca.

α × β = 43 × (-1) = = ; ca = −43 = .

Hence, product of zeroes (α × β) = ca is verified.

Hint

1.The roots of the quadratic expression of the form ax2+bx+c are given by the formula: x = −b±b2−4ac2a where α = −b+b2−4ac2a ;β = −b−b2−4ac2a

2.Sum of the zeroes (α + β) = - ba

3.Product of zeroes (α × β) = ca

(i)

(i)14, -1

Given: Sum of the zeroes (α + β) =

Product of zeroes (α × β) =

Substitute the values in quadratic Polynomial:k(x2 - (α + β) x + (α × β)) where k is any real number

Let us assume k=1.

Required Quadratic Equation = (x2- x + )

= x2-14x - 1

(ii)

(ii)√2 ,13

Given: Sum of the zeroes (α + β) = √2-√2

Product of zeroes (α × β) =

Substitute the values in quadratic Polynomial:k(x2 - (α + β) x + (α × β)) where k is any real number

Let us assume k=1.

Required Quadratic Equation = (x2- √2-√2 x + )

= x2−2x+13

(iii)

(iii) 0, 5

Given: Sum of the zeroes (α + β) =

Product of zeroes (α × β) = 5-5

Substitute the values in quadratic Polynomial:k(x2 - (α + β) x + (α × β)) where k is any real number

Let us assume k=1.

Required Quadratic Equation = (x2- x + 5-5)

= x2−5

(iv)

(iv) 1, 1

Given: Sum of the zeroes (α + β) =

Product of zeroes (α × β) =

Substitute the values in quadratic Polynomial:k(x2 - (α + β) x + (α × β)) where k is any real number

Let us assume k=1.

Required Quadratic Equation = (x2- x + )

= x2−x+1

(v)

(v)−14 ,14

Given: Sum of the zeroes (α + β) =

Product of zeroes (α × β) =

Substitute the values in quadratic Polynomial:k(x2 - (α + β) x + (α × β)) where k is any real number

Let us assume k=1.

Required Quadratic Equation = x2- () x +

= x2+14x+14

(vi)

(vi) 4,1

Given: Sum of the zeroes (α + β) =

Product of zeroes (α × β) =

Substitute the values in quadratic Polynomial:k(x2 - (α + β) x + (α × β)) where k is any real number

Let us assume k=1.

Required Quadratic Equation = x2- x +

= x2−4x+1

Hint

We know that the general equation of a quadratic polynomial is:

k(x2 - (sum of roots) x + (product of roots)) where k is any real number.