These go beyond the standard Exercise 4.1–4.3 questions into genuine HOTS (Higher Order Thinking Skills) territory — parameter-based equal-roots problems, an assertion-reason question, a reverse-engineered equation from irrational roots, a digit-based word problem, and two abstract/general reasoning questions. See the standard-level Solutions post first if you haven’t already.
Extra Questions (HOTS Level): Quadratic Equations (Class 10 Maths Chapter 4)
Q1. Find the value(s) of p for which the equation (p − 2)x² + 8x + (p + 4) = 0, p ≠ 2, has real and equal roots.
Here a = (p−2), b = 8, c = (p+4). For equal roots, D = 0:
D = 8² − 4(p−2)(p+4) = 64 − 4(p² + 2p − 8) = 64 − 4p² − 8p + 32 = 96 − 8p − 4p².
Setting D = 0: 4p² + 8p − 96 = 0 ⇒ p² + 2p − 24 = 0 ⇒ (p+6)(p−4) = 0.
Both values satisfy p ≠ 2.
p = −6 or p = 4.
Q2. Assertion (A): The quadratic equation x² + 6x + 9 = 0 has real and equal roots. Reason (R): A quadratic equation ax² + bx + c = 0 (a≠0) has real roots whenever b² ≥ 4ac. Which is correct: (a) both A and R true, R correctly explains A; (b) both A and R true, but R does NOT correctly explain A; (c) A true, R false; (d) A false, R true?
Checking A: for x²+6x+9=0, D = 36−36 = 0, so the roots are real and equal — A is true.
Checking R: the general statement “b² ≥ 4ac ⇒ real roots” is itself true — it covers both the D>0 (distinct) and D=0 (equal) cases, so R is also true.
However, R only explains why the roots are real; it does not by itself explain why they are specifically equal (that needs the stricter condition b² = 4ac). So R does not correctly explain A.
Answer: (b) — both A and R are true, but R is not the correct explanation of A.
Q3. Form a quadratic equation with rational (integer) coefficients, given that one of its roots is 3 + √7.
Since the coefficients must be rational, irrational roots of a quadratic always occur in conjugate pairs. So the other root must be 3 − √7.
Sum of roots = (3+√7) + (3−√7) = 6. Product of roots = (3+√7)(3−√7) = 9 − 7 = 2.
Using x² − (sum)x + (product) = 0:
x² − 6x + 2 = 0.
Q4. A two-digit number is such that the product of its digits is 18. When 27 is added to the number, its digits interchange their places. Find the number.
Let the ten’s digit = x and the unit’s digit = y, so the number = 10x+y and the digit product xy = 18.
Adding 27 reverses the digits: 10x + y + 27 = 10y + x ⇒ 9x − 9y = −27 ⇒ y = x + 3.
Substituting into xy = 18: x(x+3) = 18 ⇒ x² + 3x − 18 = 0 ⇒ (x+6)(x−3) = 0. Rejecting the negative root, x = 3, so y = 6.
Check: number = 36, reversed = 63, and 36 + 27 = 63. ✓
The number is 36.
Q5. If α and β are the roots of x² − px + q = 0, find, in terms of p and q, the quadratic equation whose roots are (α+1) and (β+1).
For the given equation, sum of roots α+β = p and product αβ = q.
New sum = (α+1)+(β+1) = (α+β) + 2 = p + 2.
New product = (α+1)(β+1) = αβ + (α+β) + 1 = q + p + 1.
Using x² − (new sum)x + (new product) = 0:
x² − (p+2)x + (p+q+1) = 0.
Q6. If the roots of ax² + bx + c = 0 (a≠0) are in the ratio 3 : 4, prove that 12b² = 49ac.
Let the roots be 3k and 4k for some k≠0.
Sum of roots: 3k + 4k = 7k = −b/a ⇒ k = −b/(7a).
Product of roots: (3k)(4k) = 12k² = c/a.
Substituting k: 12 × b²/(49a²) = c/a ⇒ 12b²/(49a²) = c/a ⇒ 12b² = 49a² × (c/a) = 49ac.
Hence 12b² = 49ac, proved.
How These Differ From the Textbook Exercises
Exercises 4.1–4.3 focus on recognising quadratic equations, solving them directly by factorisation, and applying the discriminant to a given numeric equation. These HOTS questions instead ask you to work backwards from conditions (build an equation from given roots, as in Q3 and Q6), reason abstractly in terms of p, q, a, b, c rather than specific numbers (Q5, Q6), and carefully distinguish between related-but-different concepts, such as “real” versus “equal” roots in the assertion-reason format (Q2) — exactly the kind of multi-step, concept-linking reasoning the CBSE board favours in its higher-mark and competency-based questions.
