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Question -

For each of thefollowing, find a quadratic polynomial whose sum and product respectively ofthe zeroes are as given. Also find the zeroes of these polynomials byfactorisation.

(i) (–8/3), 4/3

(ii) 21/8, 5/16

(iii) -2√3, -9

(iv) (-3/(2√5)), -½



Answer -

Solution:

(i) Sum of the zeroes= – 8/3

Product of the zeroes= 4/3

P(x) = x2 – (sum of the zeroes) + (product of the zeroes)

Then, P(x)= x2 – (-8x)/3 + 4/3

P(x)= 3x2 + 8x + 4

Using splitting themiddle term method,

3x2 + 8x + 4 = 0

3x2 + (6x + 2x) + 4 = 0

3x2 + 6x + 2x + 4 = 0

3x(x + 2) + 2(x + 2) =0

(x + 2)(3x + 2) = 0

 x = -2, -2/3

(ii) Sum of the zeroes= 21/8

Product of the zeroes= 5/16

P(x) = x2 – (sum of the zeroes) + (product of the zeroes)

Then, P(x)= x2 – 21x/8 + 5/16

P(x)= 16x2 – 42x + 5

Using splitting themiddle term method,

16x2 – 42x + 5 = 0

16x2 – (2x + 40x) + 5 = 0

16x2 – 2x – 40x + 5 = 0

2x (8x – 1) – 5(8x –1) = 0

(8x – 1)(2x – 5) = 0

 x = 1/8, 5/2

(iii) Sum of thezeroes = – 2√3

Product of the zeroes= – 9

P(x) = x2 – (sum of the zeroes) + (product of the zeroes)

Then, P(x) = x2 – (-2√3x) – 9

Using splitting themiddle term method,

x2 + 2√3x – 9 = 0

x2 + (3√3x – √3x) – 9 = 0

x(x + 3√3) – √3(x +3√3) = 0

(x – √3)(x + 3√3) = 0

 x =  √3,-3√3

(iv) Sum of the zeroes= -3/2√5x

Product of the zeroes= – ½

P(x) = x2 – (sum of the zeroes) + (product of the zeroes)

Then, P(x)= x2  -(-3/2√5x) – ½

P(x)= 2√5x2 + 3x – √5

Using splitting themiddle term method,

2√5x2 + 3x – √5 = 0

2√5x2 + (5x – 2x) – √5 = 0

2√5x2 – 5x + 2x – √5 = 0

√5x (2x + √5) – (2x +√5) = 0

(2x + √5)(√5x – 1) = 0

 x = 1/√5, -√5/2

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