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 Name | Class 10 Mathematics NCERT Textbook |
Chapter | Chapter 2 Polynomials |
Exercise | Ex 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) 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
Solution
(i) x^{2} – 2x – 8 = x2 -4x +4x -8
= x(x-) + 2(x-4) = (x+2)(x-4)
So, the value of p(x) = x^{2} – 2x – 8 is zero when (x+2)=0 and (x-4)=0
When x = -2 or x = . This, the zeroes of x^{2} – 2x – 8 are 4 and -2.
∴ Sum of the zeroes = (-2) + (4) =2 = -(-2)/1 = -(coefficient of x)/coefficient of x^{2}
Product of the zeroes = (-2) × (4) = -8 = (-8)/1 = constant term / coefficient of x^{2}
(ii) 4s^{2} – 4s + 1 = 4s^{2} - 2s - 2s + 1
= 2s(2s -1) - 1(2s-1)
= (2s -1) (2s-1)
So, the value of 4s^{2} -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 4s^{2} - 4s + 1 are 1/2 and 1/2.
∴ Sum of the zeroes = 1/2 + 1/2 = 1 = -(-1)/1 = -(coefficient of s)/ coefficient of s^{2}
and, product of zeroes = (1/2) (1/2) = 1/4 = Constant term/Coefficient of s^{2}
(iii) 6x^{2} – 3 – 7x = 6x^{2} = 6x^{2 }– 7x – 3
= 6x^{2 }– 9x – 2x -3
= 3x(2x -3) +1(2x-3)
= (3x + 1) (2x - 3)
The value of 6x^{2} - 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 6x^{2} - 3 - 7x are -1/3 and 3/2.
Sum of zeroes = -1/3 + 3/2 = 7/6 = -(-7)/6 = -(Coefficient of x)/Coefficient of x^{2}
Product of zeroes = -1/3 × 3/2 = -1/2 = -3/6 = Constant term/Coefficient of x^{2}.
(iv) 4u^{2} + 8u = 4u^{2} + 8u + 0
= 4u(u + 2)
The value of 4u^{2} + 8u is zero when 4u = 0 or u + 2 = 0, i.e., u = 0 or u = - 2
Therefore, the zeroes of 4u^{2} + 8u are 0 and - 2.
Sum of zeroes = 0 + (-2) = -2 = -(8)/4 = -(Coefficient of u)/Coefficient of u^{2}
Product of zeroes = 0 × (-2) = 0 = 0/4 = Constant term/Coefficient of u^{2}.
(v) t^{2} – 15
= t^{2 }- 0.t - 15
= (t - √15) (t + √15)
The value of t^{2} - 15 is zero when t - √15 = 0 or t + √15 = 0, i.e., when t = √15 or t = -√15
Therefore, the zeroes of t^{2} - 15 are √15 and -√15.Sum of zeroes = √15 + -√15 = 0 = -0/1 = -(Coefficient of t)/Coefficient of t^{2}
Product of zeroes = (√15) (-√15) = -15 = -15/1 = Constant term/Coefficient of u^{2}.
(vi) 3x^{2} – x – 4
= (3x - 4) (x + 1)
The value of 3x^{2} – 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 3x^{2} – x – 4 are 4/3 and -1.
Sum of zeroes = 4/3 + (-1) = 1/3 = -(-1)/3 = -(Coefficient of x)/Coefficient of x^{2}
Product of zeroes = 4/3 × (-1) = -4/3 = Constant term/Coefficient of x^{2}.
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 ax^{2} + 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 4x^{2} - x -4.
(ii) √2 , 1/3
Let the quadratic polynomial be ax^{2} + 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 3x^{2} -3√2x +1.
(iii) 0, √5
Let the polynomial be ax^{2} + 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 x^{2} + √5.
(iv) 1, 1
Let the polynomial be ax^{2} + 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 x^{2} - x +1.
(v) -1/4 ,1/4
Let the polynomial be ax^{2} + 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 4x^{2} + x +1.
(vi) 4,1
Let the polynomial be ax^{2} + 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 x^{2} - 4x +1.
These NCERT Solutions for Class 10 Maths will be very helpful in scoring good marks in the examinations.