Class 10 Maths Chapter 2 Extra Questions – Polynomials (HOTS)

Class 10 Maths Chapter 2 Extra Questions (HOTS Level)

These higher-order-thinking questions go beyond the standard Exercise 2.1 pattern — reverse-engineering, general proofs, assertion-reason, and fresh word problems, all with numbers not used in the NCERT textbook exercise. These Class 10 Mathematics Chapter 2 important questions are handy for last-minute exam practice.

Last Updated: September 23, 2026

How to Approach the HOTS Questions in This Chapter

1. Q1 (Reverse-engineering). If α, β are the zeroes of x² − 5x + 6, find a quadratic polynomial whose zeroes are 1/α and 1/β.
   For x²−5x+6: α+β=5, αβ=6. New sum = 1/α+1/β = (α+β)/(αβ) = 5/6. New product = 1/(αβ) = 1/6. Required polynomial: x²−(5/6)x+1/6, or clearing fractions, 6x² − 5x + 1.
2. Q2 (General/abstract proof). If α, β are the zeroes of ax²+bx+c, prove that α²+β² = (b²−2ac)/a².
   Proof: α+β = −b/a and αβ = c/a. Now α²+β² = (α+β)² − 2αβ = b²/a² − 2c/a = (b²−2ac)/a². (Check with x²−5x+6, zeroes 2,3: α²+β²=4+9=13, and (25−12)/1=13. ✓)
3. Q3 (Fresh word problem). A rectangular field has length 3 m more than twice its breadth, and its area is 90 m². Form a quadratic equation in the breadth and find the dimensions.
   Let breadth = x m, length = (2x+3) m. Area: x(2x+3) = 90 ⇒ 2x²+3x−90=0. Factorising: (x−6)(2x+15)=0 ⇒ x=6 or x=−7.5 (rejected). Breadth = 6 m, length = 15 m. Check: 6×15=90. ✓
4. Q4 (Assertion-Reason). Assertion (A): If the sum of the zeroes of a quadratic polynomial is 0, the polynomial has no x-term.
   Reason (R): For ax²+bx+c, sum of zeroes = −b/a; if this is 0, then b=0.
   Options: (a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
   Answer: (a). Sum=0 forces −b/a=0, i.e. b=0, so the polynomial reduces to ax²+c — R directly explains A.
5. Q5 (Reverse-engineering). Find the value of k such that the sum of the zeroes of x²−(k+6)x+2(2k−1) is half of the product of its zeroes.
   Sum = (k+6), Product = 2(2k−1) = 4k−2. Given: k+6 = (1/2)(4k−2) = 2k−1 ⇒ 7 = k. k = 7. Check: sum=13, product=26, half of 26 is 13. ✓
6. Q6 (Reverse-engineering). If the zeroes of x²+px+q are double the zeroes of 2x²−5x−3, find p and q.
   Zeroes of 2x²−5x−3=0: (2x+1)(x−3)=0 ⇒ x = 3 or −1/2. Doubled zeroes: 6 and −1. New sum = 5 = −p ⇒ p = −5. New product = −6 = q ⇒ q = −6. Check: x²−5x−6=(x−6)(x+1)=0 gives zeroes 6,−1. ✓

Related

• Solutions
• Extra Questions (HOTS)
• Class 10 Maths Book

Frequently Asked Questions

How would you find the zeroes of the polynomial p(x) = x squared minus 5x plus 6?
Factor the polynomial as (x-2)(x-3), giving zeroes at x=2 and x=3.

If the sum and product of the zeroes of a quadratic polynomial are 4 and 3 respectively, how would you form the polynomial?
Using the relation x squared minus sum times x plus product, the polynomial is x squared minus 4x plus 3.

Written by Satish

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