Ch 2: Polynomials (NCERT Solutions)
Based on the latest rationalized CBSE syllabus. Master graphical zeroes and the relationship between zeroes and coefficients.
Exercise 2.1 Solutions
💡 Concept Trick: The number of zeroes of a polynomial p(x) is simply the number of times its graph intersects the x-axis.
Q1. The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.
Solution:
- (i) The graph is a straight line parallel to the x-axis. It does not intersect the x-axis at any point. Therefore, the number of zeroes is 0.
- (ii) The graph intersects the x-axis at exactly one point. Therefore, the number of zeroes is 1.
- (iii) The graph intersects the x-axis at three distinct points. Therefore, the number of zeroes is 3.
- (iv) The graph intersects the x-axis at two distinct points. Therefore, the number of zeroes is 2.
- (v) The graph intersects the x-axis at four distinct points. Therefore, the number of zeroes is 4.
- (vi) The graph intersects the x-axis at one point and touches it at two points (total 3 points of contact). Therefore, the number of zeroes is 3.
Exercise 2.2 Solutions
Q1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x² - 2x - 8
To find zeroes, let p(x) = 0
x² - 4x + 2x - 8 = 0 (By splitting the middle term)
x(x - 4) + 2(x - 4) = 0
(x - 4)(x + 2) = 0
So, the zeroes are α = 4 and β = -2.
Verification:
Sum of zeroes (α + β) = 4 + (-2) = 2. Also, -b/a = -(-2)/1 = 2.
Product of zeroes (αβ) = 4 × (-2) = -8. Also, c/a = -8/1 = -8.
Verified.
(ii) 4s² - 4s + 1
Let p(s) = 0
4s² - 2s - 2s + 1 = 0
2s(2s - 1) - 1(2s - 1) = 0
(2s - 1)(2s - 1) = 0
So, the zeroes are α = 1/2 and β = 1/2.
Verification:
Sum = 1/2 + 1/2 = 1. Also, -b/a = -(-4)/4 = 1.
Product = (1/2) × (1/2) = 1/4. Also, c/a = 1/4.
Verified.
(iii) t² - 15
Let p(t) = 0
t² - 15 = 0 ⇒ t² = 15 ⇒ t = ±√15
So, the zeroes are α = √15 and β = -√15.
Verification:
Sum = √15 + (-√15) = 0. Also, -b/a = -0/1 = 0. (Since there is no 't' term, b=0)
Product = (√15) × (-√15) = -15. Also, c/a = -15/1 = -15.
Verified.
Q2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
Formula required: A quadratic polynomial is given by k[x² - (Sum)x + (Product)], where k is any non-zero real number.
(i) 1/4, -1
Sum = 1/4, Product = -1
Polynomial = x² - (1/4)x + (-1)
Multiplying by 4 to remove the fraction (k=4), we get: 4x² - x - 4
(ii) √2, 1/3
Sum = √2, Product = 1/3
Polynomial = x² - (√2)x + (1/3)
Multiplying by 3 (k=3), we get: 3x² - 3√2x + 1
(iii) 0, √5
Sum = 0, Product = √5
Polynomial = x² - (0)x + √5
Answer: x² + √5
Previous Year Questions (PYQs)
Q. If α and β are the zeroes of the polynomial x² - 5x + 6, find the value of α² + β².
Solution:
Given polynomial: p(x) = x² - 5x + 6. Here a=1, b=-5, c=6.
We know that:
Sum of zeroes (α + β) = -b/a = -(-5)/1 = 5
Product of zeroes (αβ) = c/a = 6/1 = 6
To find α² + β², we use the algebraic identity:
(α + β)² = α² + β² + 2αβ
α² + β² = (α + β)² - 2αβ
Substituting the values:
α² + β² = (5)² - 2(6)
α² + β² = 25 - 12 = 13
Therefore, the value of α² + β² is 13.