Chapter 4 — Quadratic Equations is one of the most exam-heavy chapters in the current 2026-27 CBSE Class 10 Maths syllabus, spanning three exercises (4.1, 4.2, 4.3). Below are complete, original, step-by-step solutions covering identifying quadratic equations, solving by factorisation, and determining the nature of roots.
NCERT Solutions for Class 10 Maths Chapter 4: Quadratic Equations
Exercise 4.1 — Checking Whether an Equation Is Quadratic
Q1. Check whether the following are quadratic equations:
(i) (x+1)² = 2(x−3)
Expanding: x²+2x+1 = 2x−6 → x²+2x+1−2x+6 = 0 → x²+7 = 0. Degree 2 — quadratic equation.
(ii) x²−2x = (−2)(3−x)
x²−2x = −6+2x → x²−2x−2x+6 = 0 → x²−4x+6 = 0. Quadratic equation.
(iii) (x−2)(x+1) = (x−1)(x+3)
LHS = x²−x−2, RHS = x²+2x−3. Subtracting: −3x+1 = 0. This is linear — not a quadratic equation.
(iv) (x−3)(2x+1) = x(x+5)
LHS = 2x²−5x−3, RHS = x²+5x. Subtracting: x²−10x−3 = 0. Quadratic equation.
(v) (2x−1)(x−3) = (x+5)(x−1)
LHS = 2x²−7x+3, RHS = x²+4x−5. Subtracting: x²−11x+8 = 0. Quadratic equation.
(vi) x²+3x+1 = (x−2)²
RHS = x²−4x+4. Subtracting: 7x−3 = 0. This is linear — not a quadratic equation.
(vii) (x+2)³ = 2x(x²−1)
LHS = x³+6x²+12x+8, RHS = 2x³−2x. Subtracting: −x³+6x²+14x+8 = 0. This is a cubic (degree 3) — not a quadratic equation.
(viii) x³−4x²−x+1 = (x−2)³
RHS = x³−6x²+12x−8. Subtracting: 2x²−13x+9 = 0. Degree 2 — quadratic equation.
Q2. Represent the following situations as quadratic equations:
(i) The area of a rectangular plot is 528 m². The length is one more than twice its breadth.
Let breadth = x m, so length = (2x+1) m. Area: x(2x+1) = 528 → 2x²+x−528 = 0.
(ii) The product of two consecutive positive integers is 306.
Let the integers be x and x+1. x(x+1) = 306 → x²+x−306 = 0.
(iii) Rohan’s mother is 26 years older than him. The product of their ages 3 years from now will be 360.
Let Rohan’s present age = x years, so his mother’s = (x+26). In 3 years: (x+3)(x+29) = 360 → x²+32x+87 = 360 → x²+32x−273 = 0.
(iv) A train travels 480 km at a uniform speed. If the speed had been 8 km/h less, it would have taken 3 hours more.
Let the speed = x km/h. 480/(x−8) − 480/x = 3 → 480×8 = 3x(x−8) → x²−8x−1280 = 0.
Exercise 4.2 — Solving by Factorisation
Q1. Find the roots by factorisation:
(i) x²−3x−10 = 0
Split −3x as −5x+2x: (x−5)(x+2) = 0 → x = 5 or x = −2.
(ii) 2x²+x−6 = 0
Split x as 4x−3x: 2x(x+2)−3(x+2) = 0 → (x+2)(2x−3) = 0 → x = −2 or x = 3/2.
(iii) √2 x²+7x+5√2 = 0
Split 7x as 5x+2x: x(√2x+5)+√2(√2x+5) = 0 → (√2x+5)(x+√2) = 0 → x = −5/√2 or x = −√2.
(iv) 2x²−x+1/8 = 0
Multiply by 8: 16x²−8x+1 = 0 → split −8x as −4x−4x: 4x(4x−1)−1(4x−1) = 0 → (4x−1)² = 0 → x = 1/4 (repeated root).
(v) 100x²−20x+1 = 0
Split −20x as −10x−10x: 10x(10x−1)−1(10x−1) = 0 → (10x−1)² = 0 → x = 1/10 (repeated root).
Q2. John and Jivanti together have 45 marbles. Both lost 5 marbles each, and the product of the marbles they now have is 124. Find how many marbles each had at the start.
Let John have x, so Jivanti has (45−x). After losing 5 each: (x−5)(40−x) = 124 → x²−45x+324 = 0 → (x−36)(x−9) = 0 → x = 36 or x = 9.
John had 36 and Jivanti had 9 marbles (or vice versa).
Q3. Find two numbers whose sum is 27 and product is 182.
x(27−x) = 182 → x²−27x+182 = 0 → (x−14)(x−13) = 0 → the numbers are 13 and 14.
Q4. Find two consecutive positive integers, the sum of whose squares is 365.
x²+(x+1)² = 365 → 2x²+2x−364 = 0 → x²+x−182 = 0 → (x+14)(x−13) = 0 → x = 13 (rejecting negative). The integers are 13 and 14.
Q5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Let base = x, altitude = x−7. x²+(x−7)² = 13² → 2x²−14x−120 = 0 → x²−7x−60 = 0 → (x−12)(x+5) = 0 → x = 12 (rejecting negative). Base = 12 cm, altitude = 5 cm.
Q6. A cottage industry produces pottery articles. The cost of production per article (in ₹) was 3 more than twice the number of articles produced that day. The total cost of production was ₹90. Find the number of articles and the cost of each.
Let the number of articles = x. Cost per article = 2x+3. x(2x+3) = 90 → 2x²+3x−90 = 0 → (2x+15)(x−6) = 0 → x = 6 (rejecting negative). Number of articles = 6, cost per article = ₹15.
Exercise 4.3 — Nature of Roots
Q1. Find the nature of the roots; if real, find them:
(i) 2x²−3x+5 = 0
D = b²−4ac = 9−40 = −31 < 0 → no real roots.
(ii) 3x²−4√3x+4 = 0
D = 48−48 = 0 → real and equal roots. x = −b/2a = 4√3/6 = 2√3/3 (both roots).
(iii) 2x²−6x+3 = 0
D = 36−24 = 12 > 0 → real, distinct roots. x = (6±√12)/4 = (3±√3)/2.
Q2. Find k so that the equation has two equal roots:
(i) 2x²+kx+3 = 0
D = k²−24 = 0 → k = ±2√6.
(ii) kx(x−2)+6 = 0 → kx²−2kx+6 = 0
D = 4k²−24k = 0 → 4k(k−6) = 0 → k = 0 (rejected, coefficient of x² cannot be zero) or k = 6.
Q3. Is it possible to design a rectangular mango grove whose length is twice its breadth, with area 800 m²? If so, find its length and breadth.
Let breadth = x, length = 2x. 2x² = 800 → x² = 400 → x = 20. Yes, possible: breadth = 20 m, length = 40 m.
Q4. Is this situation possible? The sum of the ages of two friends is 20 years. Four years ago, the product of their ages was 48.
Let one’s age = x, other = 20−x. Four years ago: (x−4)(16−x) = 48 → x²−20x+112 = 0. D = 400−448 = −48 < 0. Not possible — no real solution exists.
Q5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth.
l+w = 40, l×w = 400. w(40−w) = 400 → w²−40w+400 = 0. D = 1600−1600 = 0 → w = 20. Yes, possible: length = width = 20 m (a square).
Why This Chapter Matters for Boards
Quadratic Equations is a near-guaranteed 6–8 mark contributor to the CBSE Class 10 board Maths paper every year — factorisation, the discriminant, and nature-of-roots questions appear both as standalone short-answer items and embedded inside longer word-problem questions. The reasoning skills here (forming an equation from a word problem, then checking which root actually fits the real-world context) reappear directly in Class 10 Chapter 5 (Arithmetic Progressions) and across Class 11 algebra.
Frequently Asked Questions
How many exercises are there in Class 10 Maths Chapter 4?
Three — Exercise 4.1 (identifying quadratic equations), 4.2 (solving by factorisation), and 4.3 (nature of roots), matching the current 2026-27 rationalised NCERT textbook.
Is the quadratic formula method covered in this chapter?
The discriminant (used to determine the nature of roots) is covered in Exercise 4.3; the full quadratic formula x = [−b±√(b²−4ac)]/2a is summarised in our Formulas Handbook for quick reference and application in harder problems.
Extra Questions (HOTS) | Revision Notes | Formulas Handbook | Class 10 Maths Book
