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Quadratic Equations Class 10 Maths Worksheet with Answers Chapter 4

Updated:10 Mins Read

Here we are with chapter 4 of the NCERT book, i.e., Quadratic Equations. A chapter that will be with you even after class 10. Before a test, start slow, think clearly, and write steps neatly. The worksheet moves in four levels: Basic to lock key ideas, Standard to build exam-ready fluency, Advance handles multi-step applications, and HOTS sharpens your reasoning when questions look unfamiliar. Remember, the real marks come from your method, not just the final answer.

This chapter covers below topics –

  • Quadratic equation form
  • Factorisation solving method
  • Quadratic formula usage
  • Discriminant and roots
  • Word problems modelling

Class 10 Maths Worksheet – Chapter 4: Quadratic Equations

Basic

  1. Write the standard form of a quadratic equation in x.
  2. Decide whether the given equation is quadratic (Yes/No):

    • (x − 2)² + 1 = 2x − 3
    • x(x + 1) + 8 = (x + 2)(x − 2)
    • (x + 2)³ = x³ − 4
  3. Convert into standard form:
    x(2x + 3) = x² + 1
  4. Find the roots by factorisation:
    x² − 5x + 6 = 0
  5. Find the roots by factorisation:
    2x² − 7x + 3 = 0
  6. Find the discriminant (D) for:
    2x² − 4x + 3 = 0
    Then state the nature of roots.
  7. If a quadratic equation has D = 0, then how many real roots does it have?
    Answer in one line.

Standard

  1. Solve by factorisation:
    x² − 3x − 10 = 0
  2. Solve by factorisation:
    2x² + x − 6 = 0
  3. Solve by quadratic formula:
    3x² − 2x + 1/3 = 0
  4. Find the roots (if real) using discriminant first:
    2x² − 3x + 5 = 0
  5. Find the value(s) of k so that the equation has two equal roots:
    2x² + kx + 3 = 0
  6. Represent the situation as a quadratic equation:
    Area of a rectangle is 528 m², and length is 1 m more than twice the breadth.
  7. Solve:
    The product of two consecutive positive integers is 306. Find the integers.

Advance

  1. Solve by factorisation:
    6x² − x − 2 = 0
  2. Solve:
    Find two numbers whose sum is 27 and product is 182.
  3. Solve:
    Find two consecutive positive integers, sum of whose squares is 365.
  4. A right triangle has hypotenuse 13 cm.
    The altitude is 7 cm less than the base.
    Find the base and altitude.
  5. A train travels 480 km at a uniform speed.
    If the speed were 8 km/h less, it would take 3 hours more.
    Find the original speed.
  6. A cottage industry makes x articles in a day.
    Cost per article (₹) is (2x + 3).
    Total cost is ₹90.
    Find x and the cost per article.
  7. Solve by formula and state the nature of roots:
    x² + 7x − 60 = 0

HOTS

  1. Without solving fully, decide the nature of roots:
    x² + 2x + 2 = 0
    Then state whether roots are real or not.
  2. If one root of 2x² − 5x + 3 = 0 is 1, find the other root without using the formula.
  3. A quadratic equation has roots 3 and −4.
    Form the equation in standard form (leading coefficient 1).
  4. The sum of two numbers is 20.
    Four years ago, the product of their ages was 48.
    Is this situation possible? If yes, find their present ages.
  5. Is it possible to design a rectangle with perimeter 80 m and area 400 m²?
    If yes, find its dimensions.
  6. A rectangular mango grove has length twice its breadth and area 800 m².
    Is it possible? If yes, find length and breadth.
  7. A circular park has diameter 13 m.
    A pole P is on the boundary such that |AP − BP| = 7 m, where A and B are opposite ends of the diameter.
    Find AP and BP (in metres).

Answer Key

Basic – Answers

  1. Ans: ax² + bx + c = 0, where a ≠ 0.
    Hint: Highest power is 2, and a cannot be 0.
  2. Ans:

    • (x − 2)² + 1 = 2x − 3 → Yes
    • x(x + 1) + 8 = (x + 2)(x − 2) → No
    • (x + 2)³ = x³ − 4 → Yes

    Hint: Simplify first, then check if degree becomes 2.

  3. Ans: x(2x + 3) = x² + 1 ⇒ 2x² + 3x = x² + 1 ⇒ x² + 3x − 1 = 0
    Hint: Bring all terms to one side.
  4. Ans: x² − 5x + 6 = 0 ⇒ (x − 2)(x − 3) = 0 ⇒ x = 2, 3
    Hint: Find two numbers with product 6 and sum 5.
  5. Ans: 2x² − 7x + 3 = 0 ⇒ (2x − 1)(x − 3) = 0 ⇒ x = 1/2, 3
    Hint: Split middle term: −7x = −6x − x.
  6. Ans: For 2x² − 4x + 3 = 0, D = b² − 4ac = (−4)² − 4(2)(3) = 16 − 24 = −8
    Nature: no real roots
    Hint: D < 0 ⇒ roots are not real.
  7. Ans: Two equal real roots.
    Hint: D = 0 ⇒ repeated root.

Standard – Answers

  1. Ans: x² − 3x − 10 = 0 ⇒ (x − 5)(x + 2) = 0 ⇒ x = 5, −2
    Hint: Product −10, sum −3.
  2. Ans: 2x² + x − 6 = 0 ⇒ (2x − 3)(x + 2) = 0 ⇒ x = 3/2, −2
    Hint: Split: x = 4x − 3x.
  3. Ans: 3x² − 2x + 1/3 = 0
    D = (−2)² − 4(3)(1/3) = 4 − 4 = 0 ⇒ equal roots
    x = −b/(2a) = 2/(6) = 1/3
    Hint: D = 0 makes it fast.
  4. Ans: 2x² − 3x + 5 = 0
    D = (−3)² − 4(2)(5) = 9 − 40 = −31 < 0 ⇒ no real roots
    Hint: Check D before solving.
  5. Ans: For 2x² + kx + 3 = 0, equal roots ⇒ D = 0
    k² − 4(2)(3) = 0 ⇒ k² − 24 = 0 ⇒ k = ±2√6
    Hint: Equal roots ⇒ discriminant zero.
  6. Ans: Let breadth = x m, length = (2x + 1) m
    Area: x(2x + 1) = 528 ⇒ 2x² + x − 528 = 0
    Hint: Area = length × breadth.
  7. Ans: Let integers be x and x + 1
    x(x + 1) = 306 ⇒ x² + x − 306 = 0
    (x + 18)(x − 17) = 0 ⇒ x = 17 (positive)
    Integers: 17 and 18
    Hint: Consecutive ⇒ x and x + 1.

Advance – Answers

  1. Ans: 6x² − x − 2 = 0
    6x² + 3x − 4x − 2 = 0 ⇒ 3x(2x + 1) − 2(2x + 1) = 0
    (3x − 2)(2x + 1) = 0 ⇒ x = 2/3, −1/2
    Hint: Split middle term carefully.
  2. Ans: Let numbers be x and 27 − x
    x(27 − x) = 182 ⇒ x² − 27x + 182 = 0
    (x − 13)(x − 14) = 0 ⇒ x = 13 or 14
    Numbers: 13 and 14
    Hint: Sum fixed ⇒ second number is 27 − x.
  3. Ans: Let integers be x and x + 1
    x² + (x + 1)² = 365 ⇒ 2x² + 2x + 1 = 365
    2x² + 2x − 364 = 0 ⇒ x² + x − 182 = 0
    (x − 13)(x + 14) = 0 ⇒ x = 13 (positive)
    Integers: 13 and 14
    Hint: Expand (x + 1)² = x² + 2x + 1.
  4. Ans: Let base = x cm, altitude = (x − 7) cm, hypotenuse = 13 cm
    x² + (x − 7)² = 13²
    x² + x² − 14x + 49 = 169 ⇒ 2x² − 14x − 120 = 0
    x² − 7x − 60 = 0 ⇒ (x − 12)(x + 5) = 0 ⇒ x = 12
    Base = 12 cm, altitude = 5 cm
    Hint: Use Pythagoras theorem.
  5. Ans: Let speed = v km/h
    Time: 480/v and 480/(v − 8)
    480/(v − 8) = 480/v + 3
    Multiply by v(v − 8): 480v = 480(v − 8) + 3v(v − 8)
    480v = 480v − 3840 + 3v² − 24v
    3v² − 24v − 3840 = 0 ⇒ v² − 8v − 1280 = 0
    (v − 40)(v + 32) = 0 ⇒ v = 40
    Hint: Speed cannot be negative, so take v = 40.
  6. Ans: Total cost: x(2x + 3) = 90 ⇒ 2x² + 3x − 90 = 0
    (2x + 15)(x − 6) = 0 ⇒ x = 6 (positive)
    Cost per article = 2(6) + 3 = ₹15
    Hint: Articles count must be positive.
  7. Ans: x² + 7x − 60 = 0
    D = 7² − 4(1)(−60) = 49 + 240 = 289 = 17²
    x = (−7 ± 17)/2 ⇒ x = 5, −12
    Nature: two distinct real roots
    Hint: Perfect-square discriminant makes roots neat.

HOTS – Answers

  1. Ans: For x² + 2x + 2 = 0, D = 2² − 4(1)(2) = 4 − 8 = −4 < 0
    So roots are not real
    Hint: D < 0 ⇒ no real roots.
  2. Ans: For 2x² − 5x + 3 = 0, product of roots = c/a = 3/2
    If one root is 1, other root = (3/2) ÷ 1 = 3/2
    Hint: Use product of roots = c/a.
  3. Ans: Roots 3 and −4 ⇒ equation: (x − 3)(x + 4) = 0
    x² + x − 12 = 0
    Hint: (x − α)(x − β) = 0.
  4. Ans: Let present ages be x and 20 − x
    Four years ago: (x − 4)(16 − x) = 48
    −x² + 20x − 64 = 48 ⇒ x² − 20x + 112 = 0
    D = 400 − 448 = −48 < 0 ⇒ not possible (no real ages)
    Hint: If D < 0, situation is impossible.
  5. Ans: Let length = l, breadth = b
    Perimeter 80 ⇒ l + b = 40
    Area 400 ⇒ lb = 400
    So t² − 40t + 400 = 0 (where t is l or b)
    D = 1600 − 1600 = 0 ⇒ possible, equal sides
    l = b = 20 m
    Hint: Use l = 40 − b and substitute.
  6. Ans: Let breadth = x m, length = 2x m
    Area: 2x² = 800 ⇒ x² = 400 ⇒ x = 20
    Breadth = 20 m, length = 40 m
    Hint: x must be positive.
  7. Ans: Let BP = x, then AP = x + 7, and AB = 13 (diameter)
    ∠APB = 90°, so AP² + BP² = AB²
    (x + 7)² + x² = 13²
    2x² + 14x + 49 = 169 ⇒ 2x² + 14x − 120 = 0
    x² + 7x − 60 = 0 ⇒ (x + 12)(x − 5) = 0 ⇒ x = 5 (positive)
    BP = 5 m, AP = 12 m
    Hint: Use Pythagoras because angle in semicircle is 90°.

Worksheet for Other chapters

Amit, a BE in Mechanical Engineering, is a math enthusiast dedicated to making math fun and accessible for kids in classes 1 to 10. With a knack for simplifying complex concepts, Amit offers easy-to-understand solutions, fostering a love for math in young minds across India.

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