2. Polynomials Mathematics Exercise - 2.2 class 10 Maths in English - CBSE Study
NCERT Solutions for Class 10 Mathematics are carefully prepared according to the latest CBSE syllabus and NCERT textbooks to help students understand every concept clearly. These solutions cover all important 2. Polynomials with detailed explanations and step-by-step answers for better exam preparation. Each Exercise 2.2 is explained in simple language so that students can easily grasp the fundamentals and improve their academic performance. The study material is designed to support daily homework, revision practice, and final exam preparation for Class 10 students. With accurate answers, concept clarity, and structured content, these NCERT solutions help learners build confidence and score higher marks in their examinations. Whether you are revising a specific topic or preparing an entire chapter, this resource provides reliable and syllabus-based guidance for complete success in Mathematics.
Class 10 English Medium Mathematics All Chapters:
2. Polynomials
2. Exercise 2.2
Chapter 2. Polynomials
Exercise 2.2
Q1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x2 – 2x – 8
Solution:
Using Middle term splitting method:
x2 – 4x + 2x – 8 = 0
⇒ x(x – 4) + 2(x – 4) = 0
⇒ x(x – 4) + 2(x – 4) = 0
⇒ (x – 4) (x + 2) = 0
⇒ x – 4 = 0 x + 2 = 0
⇒ x = 4, x = – 2
Zeroes; α = 4 β = – 2
Verifying the relationship between the zeroes and the coefficients.
a = 1, b = – 2, and c = – 8
(ii) 4s2 – 4s + 1
Solution:
4s2– 4s + 1 = 0
⇒ 4s2 – 2s – 2s + 1 = 0
⇒ 2s (2s – 1) –1(2s – 1) = 0
⇒ (2s – 1) (2s – 1) = 0
⇒ 2s – 1= 0, 2s – 1= 0
⇒ 2s = 1, 2s = 1
Coefficients
a = 4, b = – 4, c = 1
Relationship between zeroes and coefficients
L.H.S = R.H.S
L.H.S = R.H.S
Hence verified the relationship between coefficients and the zeroes in both cases.
(iii) 6x2 – 3 – 7x
Solution:
⇒ 6x2 – 3 – 7x = 0
⇒ 6x2 – 7x – 3 = 0 (After rearranging the equation)
⇒ 6x2 – 9x + 2x – 3 = 0
⇒ 3x (2x – 3) +1 (2x – 3) = 0
⇒ (2x – 3) (3x + 1) = 0
⇒ 2x – 3 = 0, 3x + 1 = 0
⇒ 2x = 3, 3x = - 1
Hence verified the relationship between coefficients and the zeroes in both cases.
(iv) 4u2 + 8u
Solution: 4u(u + 2) = 0
4u = 0, u + 2 = 0
u = 0, u = – 2
Zeroes: a = 0, b = – 2
Coefficients:
a = 4, b = 8, c = 0
Verifying relationship between zeroes and coefficients
⇒ 0 = 0
⇒ L.H.S = R.H.S
Hence verified, the relationship between coefficients and zeroes in both cases.
(v) t2 – 15
Solution:
t2 – 15 = 0
t2 = 15
t = ±√15
t = √15, t = – √15
Zeroes: a = √15, b = – √15
Coefficients a = 1, b = 0, c = – 15
Verifying relationship between zeroes and Coefficients
Hence verified, the relationship between coefficients and zeroes in both cases.
(vi) 3x2 – x – 4
Solution:
3x2 – x – 4 = 0
⇒ 3x2 + 3x – 4x – 4 = 0
⇒ 3x (x + 1) – 4 (x + 1) = 0
⇒ (x + 1) (3x – 4) = 0
⇒ x + 1 = 0, 3x – 4 = 0
⇒ x = –1 and 3x = 4
L.H.S= R.H.S
Hence verified, the relationship between coefficients and zeroes in both cases.
Q2. . Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
Hence, required quadratic polynomial is 4x2 – x – 4
Hence, required quadratic polynomial is 3x2 –3 x + 1
Hence, required quadratic polynomial is x2 + √5
Hence, required quadratic polynomial is x2 – x + 1
Hence, required quadratic polynomial is 4X2 + x + 1
Hence, required quadratic polynomial is x2 – 4 + 1
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