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*Polynomials*

*NCERT Solutions Ex 2.2*(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

Answer:

**(i) x**= (x - 4)(x +2)

^{2}-2x-8⇒x

^{2}-2x-8 = 0 when x - 4= 0 or x + 2 = 0

∴ Zeroes of x

^{2}- 2x - 8 are 4 and -2.

Sum of zeroes =

Product of zeroes =

Hence relationship between zeroes and coefficients are verified.

**(ii) 4s**

^{2}-4s+1= (2s -1 )

^{2}

⇒ 4s

^{2}-4s+1 = 0 when 2s -1 = 0

⇒ s = 1/2

∴ Zeroes of 4s

^{2}-4s+1 are ½ and ½.

Sum of zeroes =

Product of zeroes =

Hence relationship between zeroes and coefficients are verified.

**(iii) 6x**

^{2}-3-7x= (3x + 1)(2x - 3)

⇒ 6x

^{2}-3-7x = 0 when 3x + 1 = 0 OR 2x - 3 = 0

⇒ x = -1/3 or x = 3/2 ∴ Zeroes of 6x

^{2}-3-7x are -1/3 and 3/2

Sum of zeroes =

Product of zeroes =

Hence relationship between zeroes and coefficients are verified.

**(iv) 4u**

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

⇒ 4u

^{2}+8u = 0 when 4u = 0 OR u + 2 = 0

⇒ u = 0 OR u = -2

∴ Zeroes of 4u

^{2}+8u are 0 OR -2

Sum of zeroes =

Product of zeroes =

Hence relationship between zeroes and coefficients are verified.

**(v) t**

^{2}-15= t

^{2}- 0t -15 = (t - √15)(t + √15)

⇒ The value of t

^{2}-15 is 0 when t - √15 = 0 OR t + √15 = 0

⇒ t = √15 or t = -√15.

Sum of zeroes = √15 -√15 = 0

Product of zeores = (√15) ☓ (-√15) = -15

Hence relationship between zeroes and coefficients are verified.

**(vi) 3x**

^{2}- x -4= 3x

^{2}- 3x + 4x -4 = 3x(x + 1) - 4(x + 1) = (x + 1)(3x - 4)

⇒ The value of 3x

^{2}- x -4 is 0, when x + 1 = 0 OR 3x -4 = 0

⇒ x = -1 OR x = 4/3

Sum of zeroes = -1 + 4/3 = -1/3

Product of zeroes = -1 ☓ 4/3 = -4/3

Hence relationship between zeroes and coefficients are verified.

**Q2: Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.**

(i) 1/4, -1

(ii) √2, 1/3

(iii) 0, √5

(iv) 1,1

(v)-1/4, 1/4

(vi) 4,1

Answer:

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