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I am trying to find NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.2. Give Class 10 Mathematics NCERT Solutions to complete hometask.

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Given Class 10 NCERT Solutions for Chapter 2 Polynomials Exercise 2.2 can be used to complete hometask and solve difficult questions. Through the help of CBSE NCERT Solutions for Class 10, you can able to solve the problems efficiently.

Book NameClass 10 Mathematics NCERT Textbook
ChapterChapter 2 Polynomials
ExerciseEx 2.2

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.2

1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x2 – 2x – 8
(ii) 4s2 – 4s + 1
(iii) 6x2 – 3 – 7x
(iv) 4u2 + 8u
(v) t2 – 15
(vi) 3x2 – x – 4

Solution

(i) x2 – 2x – 8 = x2 -4x +4x -8

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

So, the value of p(x) = x2 – 2x – 8 is zero when (x+2)=0 and (x-4)=0

When x = -2 or x = . This, the zeroes of x2 – 2x – 8 are 4 and -2.

∴ Sum of the zeroes = (-2) + (4) =2 = -(-2)/1 = -(coefficient of x)/coefficient of x2 

Product of the zeroes = (-2) × (4) = -8 = (-8)/1 = constant term / coefficient of x2 

(ii) 4s2 – 4s + 1 = 4s2 - 2s - 2s + 1

 = 2s(2s -1) - 1(2s-1)

= (2s -1) (2s-1)

So, the value of 4s2 -2s +1 is zero when either (2s-1)=0

or (2s -1) = 0, when x = 1/2 or x =1/2.

Thus, the zeroes of 4s2 - 4s + 1 are 1/2 and 1/2.

∴ Sum of the zeroes = 1/2 + 1/2 = 1 = -(-1)/1 = -(coefficient of s)/ coefficient of s2 

and, product of zeroes = (1/2) (1/2) = 1/4 = Constant term/Coefficient of s2 

(iii) 6x2 – 3 – 7x = 6x2 6x– 7– 3

6x– 9– 2x -3

= 3x(2x -3) +1(2x-3)

= (3x + 1) (2x - 3)

The value of 6x2 - 3 - 7x is zero when 3x + 1 = 0 or 2x - 3 = 0, i.e., x = -1/3 or x = 3/2

Therefore, the zeroes of 6x2 - 3 - 7x are -1/3 and 3/2.

Sum of zeroes = -1/3 + 3/2 = 7/6 = -(-7)/6 = -(Coefficient of x)/Coefficient of x2
Product of zeroes = -1/3 × 3/2 = -1/2 = -3/6 = Constant term/Coefficient of x2.

(iv) 4u2 + 8u = 4u2 + 8u + 0
= 4u(u + 2)

The value of 4u2 + 8u is zero when 4u = 0 or u + 2 = 0, i.e., u = 0 or u = - 2
Therefore, the zeroes of 4u2 + 8u are 0 and - 2.
Sum of zeroes = 0 + (-2) = -2 = -(8)/4 = -(Coefficient of u)/Coefficient of u2
Product of zeroes = 0 × (-2) = 0 = 0/4 = Constant term/Coefficient of u2.
 

(v) t2 – 15
= t- 0.t - 15
= (t - √15) (t + √15)
The value of t2 - 15 is zero when t - √15 = 0 or t + √15 = 0, i.e., when t = √15 or t = -√15
Therefore, the zeroes of t2 - 15 are √15 and -√15.Sum of zeroes = √15 + -√15 = 0 = -0/1 = -(Coefficient of t)/Coefficient of t2
Product of zeroes = (√15) (-√15) = -15 = -15/1 = Constant term/Coefficient of u2.

(vi) 3x2 – x – 4
= (3x - 4) (x + 1)
The value of 3x2 – x – 4 is zero when 3x - 4 = 0 and x + 1 = 0,i.e., when x = 4/3 or x = -1
Therefore, the zeroes of 3x2 – x – 4 are 4/3 and -1.
Sum of zeroes = 4/3 + (-1) = 1/3 = -(-1)/3 = -(Coefficient of x)/Coefficient of x2

Product of zeroes = 4/3 × (-1) = -4/3 = Constant term/Coefficient of x2.

2. 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

Solution

(i) 1/4 , -1

Let the quadratic polynomial be ax2 + bx + c and its zeroes α and β.

α + ß = 1/4 = -b/a
αß = -1 = -4/4 = c/a
If a = 4, then b = -1, c = -4
Therefore, the quadratic polynomial is 4x2 - x -4.

(ii) √2 , 1/3 

Let the quadratic polynomial be ax2 + bx + c and its zeroes α and β.

α + ß = √2 = 3√2/3 = -b/a
αß = 1/3 = c/a
If a = 3, then b = -3√2, c = 1
Therefore, the quadratic polynomial is 3x2 -3√2x +1.

(iii) 0, √5

Let the polynomial be ax2 + bx + c, and its zeroes be α and ß
α + ß = 0 = 0/1 = -b/a
αß = √5 = √5/1 = c/a
If a = 1, then b = 0, c = √5
Therefore, the quadratic polynomial is x2 + √5.
 

(iv) 1, 1
Let the polynomial be ax2 + bx + c, and its zeroes be α and ß
α + ß = 1 = 1/1 = -b/a
αß = 1 = 1/1 = c/a
If a = 1, then b = -1, c = 1
Therefore, the quadratic polynomial is x2 - x +1.

(v) -1/4 ,1/4
Let the polynomial be ax2 + bx + c, and its zeroes be α and ß
α + ß = -1/4 = -b/a
αß = 1/4 = c/a
If a = 4, then b = 1, c = 1
Therefore, the quadratic polynomial is 4x2 + x +1.

(vi) 4,1
Let the polynomial be ax2 + bx + c, and its zeroes be α and ß
α + ß = 4 = 4/1 = -b/a
αß = 1 = 1/1 = c/a
If a = 1, then b = -4, c = 1
Therefore, the quadratic polynomial is x2 - 4x +1.

These NCERT Solutions for Class 10 Maths will be very helpful in scoring good marks in the examinations.

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