第10类NCERT解决方案–第2章多项式–练习2.2

问题1.找到以下二次多项式的零点，并验证零点与系数之间的关系。

(i)x ² – 2x – 8

x² – 2x – 8 = x² – 4x + 2x – 8 

= x (x – 4) + 2(x – 4) 

= (x – 4) (x + 2)

Therefore, zeroes of equation x² – 2x – 8 are (4, -2)

Sum of zeroes is equal to [4 – 2]= 2 = -(-2)/1 

i.e. = -(Coefficient of x) / (Coefficient of x²)

Product of zeroes is equal to 4 × (-2) = -8 =-(8)/1  

i.e.= (Constant term) / (Coefficient of x²)

(ii)4s2 – 4s +1

4s² – 4s + 1 = 4s² – 2s – 2s +1 

= 2s(2s – 1) – 1(2s – 1) 

= (2s – 1) (2s – 1)

Therefore, zeroes of  equation 4s² – 4s +1 are (1/2, 1/2)

Sum of zeroes is equal to [(1/2) + (1/2)] = 1 = -4/4 

i.e.= -(Coefficient of s) / (Coefficient of s²)

Product of zeros is equal to [(1/2) × (1/2)] = 1/4 

i.e.= (Constant term) / (Coefficient of s² )

(iii)6x ² – 3 – 7x

6x² – 3 – 7x = 6x² – 7x – 3 

= 6x² – 9x + 2x – 3 

= 3x(2x – 3) + 1(2x – 3) 

= (3x + 1) (2x – 3)

Therefore, zeroes of equation 6x² – 3 – 7x are (-1/3, 3/2)

Sum of zeroes is equal to -(1/3) + (3/2) = (7/6) 

i.e.= -(Coefficient of x) / (Coefficient of x²)

Product of zeroes is equal to -(1/3) × (3/2) = -(3/6) 

i.e.= (Constant term) / (Coefficient of x² )

(iv)4u ² + 8u

4u² + 8u = 4u(u + 2)

Therefore, zeroes of equation 4u² + 8u are (0, -2).

Sum of zeroes is equal to [0 + (-2)] = -2 = -(8/4)

i.e. = -(Coefficient of u) / (Coefficient of u²)

Product of zeroes is equal to 0 × -2 = 0 = 0/4 

i.e. = (Constant term) / (Coefficient of u² )

(v)t ² – 15

 t² – 15

⇒ t² = 15 or t = ±√15

Therefore, zeroes of equation t² – 15 are (√15, -√15)

Sum of zeroes is equal to [√15 + (-√15)] = 0 = -(0/1) 

i.e.= -(Coefficient of t) / (Coefficient of t2)

Product of zeroes is equal to √15 × (-√15) = -15 = -15/1 

i.e. = (Constant term) / (Coefficient of t2 )

(vi)3x ² – x – 4

3x² – x – 4 = 3x² – 4x + 3x – 4

= x(3x – 4) + 1(3x – 4) 

= (3x – 4) (x + 1)

Therefore, zeroes of equation 3x² – x – 4 are (4/3, -1)

Sum of zeroes is equal to (4/3) + (-1) = (1/3) = -(-1/3) 

i.e. = -(Coefficient of x) / (Coefficient of x²)

Product of zeroes is equal to (4/3) × (-1) = (-4/3) 

i.e. = (Constant term) / (Coefficient of x²)

问题2。找到一个二次多项式，每个二次多项式分别具有给定的数字和其零的和。

(i)1/4，-1

Let two zeroes be α, β

∴ Sum of zeroes = α + β

∴ Product of zeroes = αβ

Given, Sum of zeroes = α + β = 1/4

Product of zeroes = α β = -1

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x² – (α + β)x + αβ = 0

x² – (1/4)x +(-1) = 0

4x² – x – 4 = 0

∴ 4x² – x – 4 is the quadratic polynomial.

(ii)√2，1 / 3

Let two zeroes be α, β

∴ Sum of zeroes = α + β

∴ Product of zeroes = αβ

Given Sum of zeroes = α + β =√2

Product of zeroes = αβ = 1/3

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:

x² – (α + β)x + αβ = 0

x² – (√2)x + (1/3) = 0

3x² – 3√2x + 1 = 0

∴ 3x² – 3√2x + 1 is the quadratic polynomial.

(iii)0，√5

Let two zeroes be α, β

∴ Sum of zeroes = α + β

∴ Product of zeroes = αβ

Given, Sum of zeroes = α + β = 0

Product of zeroes = αβ = √5

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x² – (α + β)x + αβ = 0

x² – (0)x + √5 = 0

∴ x² + √5 is the quadratic polynomial.

(iv)1、1

Let two zeroes be α, β

∴ Sum of zeroes = α + β

∴ Product of zeroes = αβ

Given, Sum of zeroes = α + β = 1

Product of zeroes = αβ = 1

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x² – (α + β)x + αβ = 0

x² – x + 1 = 0

∴ x² – x + 1 is the quadratic polynomial.

(v)-1 / 4、1 / 4

Let two zeroes be α, β

∴ Sum of zeroes = α + β

∴ Product of zeroes = αβ

Given, Sum of zeroes = α + β = -1/4

Product of zeroes = α β = 1/4

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x² – (α + β)x + αβ = 0

x² – (-1/4)x + (1/4) = 0

4x² + x + 1 = 0

∴ 4x² + x + 1 is the quadratic polynomial.

(vi)4，1

Let two zeroes be α, β

∴ Sum of zeroes = α + β

∴ Product of zeroes = αβ

Given, Sum of zeroes = α + β = 4

Product of zeroes = αβ = 1

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:

x² – (α + β)x + αβ = 0

x² – 4x + 1 = 0

∴ x² – 4x + 1 is the quadratic polynomial.