Important Questions Class 10 Maths Chapter 4: Quadratic Equations With Solutions

A quadratic equation is an equation in one variable where the highest power of the variable is 2.
It is written as ax² + bx + c = 0, where a, b, and c are real numbers and a ≠ 0.

Quadratic equations turn real situations into ax² + bx + c = 0, from rectangular plots to speed and distance problems. CBSE 2026 tests Important Questions Class 10 Maths Chapter 4 through standard form, roots by factorisation, nature of roots, discriminant, and word problems. The quadratic formula appears in the roots section, but completing the square is not a separate exercise in the current NCERT Reprint 2026-27. These questions cover MCQs, short answers, application problems, and previous year-style questions.

Key Takeaways

• Arithmetic Progressions: Chapter 5 focuses on identifying APs, nth term, sum of n terms, and word problems.
• Main Formula: an = a + (n - 1)d gives the term at any required position.
• Exam Trap: The nth term formula uses n - 1, not n.
• NCERT Update: Exercise 5.4 is optional and not from the examination view in NCERT.

Important Questions Class 10 Maths Chapter 4 Structure 2026

Important Questions Class 10 Maths Chapter 4: Key Formulas

Every question in this chapter uses standard form, factorisation, discriminant, or the quadratic formula. Write the equation in standard form before choosing the method.

Q1. What Is The Standard Form Of A Quadratic Equation?

The standard form of a quadratic equation is ax² + bx + c = 0. Here, a, b, and c are real numbers, and a ≠ 0.

1. Given Data:
   General quadratic equation
2. Formula Used:
   ax² + bx + c = 0
3. Key Condition:
   a ≠ 0
4. Final Result:
   Standard form = ax² + bx + c = 0

Q2. What Is The Quadratic Formula Class 10?

The quadratic formula is x = [-b ± √(b² - 4ac)] / 2a. Use it when factorisation is difficult or roots involve surds.

1. Given Data:
   ax² + bx + c = 0
2. Formula Used:
   x = [-b ± √(b² - 4ac)] / 2a
3. Final Result:
   Roots are found using the quadratic formula

Q3. What Is The Discriminant Class 10 Quadratic Equations Formula?

The discriminant is D = b² - 4ac. It decides the nature of roots without solving the full equation.

1. Given Data:
   ax² + bx + c = 0
2. Formula Used:
   D = b² - 4ac
3. Final Result:
   Discriminant = b² - 4ac

Q4. What Is The Nature Of Roots Discriminant Class 10 Rule?

The discriminant decides whether roots are distinct, equal, or not real. Check whether D is positive, zero, or negative.

1. If D > 0:
   Two distinct real roots
2. If D = 0:
   Two equal real roots
3. If D < 0:
   No real roots
4. Final Result:
   The nature of roots depends on D

Quadratic Equations Class 10 MCQ With Answers

These quadratic equations class 10 MCQ questions test standard form, discriminant, root conditions, and formula use. Each question follows the CBSE 2026 board pattern.

Q1. Which Of The Following Is A Quadratic Equation After Simplification?

The correct answer is (C) x(2x + 3) = x² + 1. After simplification, it gives x² + 3x - 1 = 0.

1. Options:
   (A) x³ + 2x = 0
   (B) x(x + 1) + 8 = (x + 2)(x - 2)
   (C) x(2x + 3) = x² + 1
   (D) 3x + 7 = 0
2. Calculation:
   x(2x + 3) = x² + 1
   2x² + 3x = x² + 1
   x² + 3x - 1 = 0
3. Final Result:
   Answer: (C) x(2x + 3) = x² + 1

Q2. What Is The Discriminant Of 2x² - 4x + 3 = 0?

The discriminant is -8. Since D < 0, the equation has no real roots.

1. Given Data:
   a = 2
   b = -4
   c = 3
2. Formula Used:
   D = b² - 4ac
3. Calculation:
   D = (-4)² - 4(2)(3)
   D = 16 - 24
   D = -8
4. Final Result:
   Answer: (B) -8

Q3. What Is The Nature Of Roots Of 3x² - 4√3x + 4 = 0?

The roots are real and equal. The discriminant equals zero.

1. Given Data:
   a = 3
   b = -4√3
   c = 4
2. Formula Used:
   D = b² - 4ac
3. Calculation:
   D = (-4√3)² - 4(3)(4)
   D = 48 - 48
   D = 0
4. Final Result:
   Answer: (B) Real and equal

Q4. If One Root Of 6x² - x - k = 0 Is 2/3, Find k.

The value of k is 2. Substitute x = 2/3 in the equation.

1. Given Data:
   Equation = 6x² - x - k = 0
   Root = 2/3
2. Formula Used:
   Substitute the root in the equation.
3. Calculation:
   6(2/3)² - 2/3 - k = 0
   6(4/9) - 2/3 - k = 0
   8/3 - 2/3 - k = 0
   k = 2
4. Final Result:
   Answer: (B) 2

Q5. Find k If 2x² + kx + 3 = 0 Has Two Equal Roots.

The value of k is ±2√6. Equal roots occur when D = 0.

1. Given Data:
   a = 2
   b = k
   c = 3
2. Formula Used:
   D = 0
3. Calculation:
   k² - 4(2)(3) = 0
   k² - 24 = 0
   k² = 24
   k = ±2√6
4. Final Result:
   Answer: (B) ±2√6

Q6. How Many Real Roots Does x² + 5 = 0 Have?

The equation has 0 real roots. Its discriminant is negative.

1. Given Data:
   x² + 5 = 0
2. Formula Used:
   D = b² - 4ac
3. Calculation:
   a = 1, b = 0, c = 5
   D = 0² - 4(1)(5)
   D = -20
4. Final Result:
   Answer: (A) 0

Q7. Which Equation Has Two Distinct Real Roots?

The equation 2x² - 7x + 6 = 0 has two distinct real roots. Its discriminant is positive.

1. Given Data:
   Equation = 2x² - 7x + 6 = 0
2. Formula Used:
   D = b² - 4ac
3. Calculation:
   D = (-7)² - 4(2)(6)
   D = 49 - 48
   D = 1
4. Final Result:
   Answer: (C) 2x² - 7x + 6 = 0

Q8. If The Roots Of ax² + bx + c = 0 Are Equal, What Is The Condition?

The condition is b² = 4ac. Equal roots occur when D = 0.

1. Given Data:
   Roots are equal.
2. Formula Used:
   D = 0
3. Calculation:
   b² - 4ac = 0
   b² = 4ac
4. Final Result:
   Answer: (B) b² = 4ac

Q9. What Is The Product Of Roots Of 3x² - x - 4 = 0?

The product of roots is -4/3. Use product = c/a.

1. Given Data:
   a = 3
   b = -1
   c = -4
2. Formula Used:
   Product of roots = c/a
3. Calculation:
   Product = -4/3
4. Final Result:
   Answer: (B) -4/3

Q10. What Is The Sum Of Roots Of 2x² - 9x + 4 = 0?

The sum of roots is 9/2. Use sum = -b/a.

1. Given Data:
   a = 2
   b = -9
   c = 4
2. Formula Used:
   Sum of roots = -b/a
3. Calculation:
   Sum = -(-9)/2
   Sum = 9/2
4. Final Result:
   Answer: (A) 9/2

Factorisation Method Quadratic Equations Class 10 Questions

These factorisation method quadratic equations class 10 questions require splitting the middle term. Set each factor equal to zero to find the roots.

Q1. Find The Roots Of x² - 3x - 10 = 0 By Factorisation.

The roots are 5 and -2. Split the middle term into -5x and 2x.

1. Given Data:
   x² - 3x - 10 = 0
2. Formula Used:
   Factorise and set each factor to zero.
3. Calculation:
   x² - 3x - 10 = x² - 5x + 2x - 10
   = x(x - 5) + 2(x - 5)
   = (x - 5)(x + 2)
4. Final Result:
   x = 5, -2

Q2. Find The Roots Of 2x² + x - 6 = 0 By Factorisation.

The roots are 3/2 and -2. Split x as 4x - 3x.

1. Given Data:
   2x² + x - 6 = 0
2. Formula Used:
   Factorise and set each factor to zero.
3. Calculation:
   2x² + x - 6 = 2x² + 4x - 3x - 6
   = 2x(x + 2) - 3(x + 2)
   = (2x - 3)(x + 2)
4. Final Result:
   x = 3/2, -2

Q3. Find The Roots Of 2x² - 5x + 3 = 0 By Factorisation.

The roots are 3/2 and 1. Split -5x as -2x - 3x.

1. Given Data:
   2x² - 5x + 3 = 0
2. Formula Used:
   Factorise and set each factor to zero.
3. Calculation:
   2x² - 5x + 3 = 2x² - 2x - 3x + 3
   = 2x(x - 1) - 3(x - 1)
   = (2x - 3)(x - 1)
4. Final Result:
   x = 3/2, 1

Q4. Find The Roots Of 100x² - 20x + 1 = 0 By Factorisation.

The roots are 1/10 and 1/10. This equation is a perfect square.

1. Given Data:
   100x² - 20x + 1 = 0
2. Formula Used:
   Perfect square identity
3. Calculation:
   100x² - 20x + 1 = (10x - 1)²
   10x - 1 = 0
   x = 1/10
4. Final Result:
   x = 1/10, 1/10

Q5. Find Two Numbers Whose Sum Is 27 And Product Is 182.

The numbers are 13 and 14. Form a quadratic equation using one variable.

1. Given Data:
   Sum = 27
   Product = 182
2. Equation Formation:
   Let one number = x
   Other number = 27 - x
3. Calculation:
   x(27 - x) = 182
   27x - x² = 182
   x² - 27x + 182 = 0
   (x - 13)(x - 14) = 0
4. Final Result:
   The numbers are 13 and 14

Q6. Find Two Consecutive Positive Integers Whose Squares Add To 365.

The integers are 13 and 14. Let the numbers be x and x + 1.

1. Given Data:
   Sum of squares = 365
2. Equation Formation:
   x² + (x + 1)² = 365
3. Calculation:
   2x² + 2x + 1 = 365
   2x² + 2x - 364 = 0
   x² + x - 182 = 0
   (x - 13)(x + 14) = 0
4. Final Result:
   The integers are 13 and 14

Q7. The Altitude Of A Right Triangle Is 7 cm Less Than Its Base. The Hypotenuse Is 13 cm. Find The Other Sides.

The base is 12 cm, and the altitude is 5 cm. Use Pythagoras theorem.

1. Given Data:
   Base = x cm
   Altitude = x - 7 cm
   Hypotenuse = 13 cm
2. Formula Used:
   Base² + Altitude² = Hypotenuse²
3. Calculation:
   x² + (x - 7)² = 13²
   x² + x² - 14x + 49 = 169
   2x² - 14x - 120 = 0
   x² - 7x - 60 = 0
   (x - 12)(x + 5) = 0
4. Final Result:
   Base = 12 cm, altitude = 5 cm

Nature Of Roots Class 10 Important Questions Using Discriminant

These nature of roots class 10 important questions use D = b² - 4ac. The discriminant tells whether roots are distinct, equal, or not real.

Q1. Find The Discriminant Of 2x² - 4x + 3 = 0 And State The Nature Of Roots.

The discriminant is -8, so the equation has no real roots.

1. Given Data:
   a = 2
   b = -4
   c = 3
2. Formula Used:
   D = b² - 4ac
3. Calculation:
   D = (-4)² - 4(2)(3)
   D = 16 - 24
   D = -8
4. Final Result:
   No real roots

Q2. Find The Discriminant Of 3x² - 2x + 1/3 = 0 And Find The Roots.

The roots are 1/3 and 1/3. The discriminant equals zero.

1. Given Data:
   a = 3
   b = -2
   c = 1/3
2. Formula Used:
   D = b² - 4ac
3. Calculation:
   D = (-2)² - 4(3)(1/3)
   D = 4 - 4
   D = 0
4. Root:
   x = -b/2a
   x = 2/6
   x = 1/3
5. Final Result:
   Roots are 1/3 and 1/3

Q3. Find The Nature Of Roots Of 2x² - 3x + 5 = 0.

The equation has no real roots. Its discriminant is negative.

1. Given Data:
   a = 2
   b = -3
   c = 5
2. Formula Used:
   D = b² - 4ac
3. Calculation:
   D = (-3)² - 4(2)(5)
   D = 9 - 40
   D = -31
4. Final Result:
   No real roots

Q4. Find The Nature Of Roots Of 3x² - 4√3x + 4 = 0. If Real, Find Them.

The roots are 2√3/3 and 2√3/3. Since D = 0, the roots are equal.

1. Given Data:
   a = 3
   b = -4√3
   c = 4
2. Formula Used:
   D = b² - 4ac
3. Calculation:
   D = (-4√3)² - 4(3)(4)
   D = 48 - 48
   D = 0
4. Root:
   x = -b/2a
   x = 4√3/6
   x = 2√3/3
5. Final Result:
   Roots are 2√3/3 and 2√3/3

Q5. Find The Nature Of Roots Of 2x² - 6x + 3 = 0. If Real, Find Them.

The roots are (3 + √3)/2 and (3 - √3)/2. Since D > 0, the roots are distinct and real.

1. Given Data:
   a = 2
   b = -6
   c = 3
2. Formula Used:
   D = b² - 4ac
3. Calculation:
   D = (-6)² - 4(2)(3)
   D = 36 - 24
   D = 12
4. Quadratic Formula:
   x = [6 ± √12]/4
   x = [6 ± 2√3]/4
   x = (3 ± √3)/2
5. Final Result:
   Roots are (3 + √3)/2 and (3 - √3)/2

Q6. Find k For Which 2x² + kx + 3 = 0 Has Two Equal Roots.

The value of k is ±2√6. Equal roots occur when D = 0.

1. Given Data:
   a = 2
   b = k
   c = 3
2. Formula Used:
   D = 0
3. Calculation:
   k² - 4(2)(3) = 0
   k² - 24 = 0
   k² = 24
4. Final Result:
   k = ±2√6

Q7. Find k For Which kx(x - 2) + 6 = 0 Has Two Equal Roots.

The value of k is 6. Reject k = 0 because the equation would not remain quadratic.

1. Given Data:
   kx(x - 2) + 6 = 0
2. Standard Form:
   kx² - 2kx + 6 = 0
3. Formula Used:
   D = 0
4. Calculation:
   (-2k)² - 4(k)(6) = 0
   4k² - 24k = 0
   4k(k - 6) = 0
   k = 0 or k = 6
5. Final Result:
   k = 6

Word Problems On Quadratic Equations Class 10 With Solutions

These word problems on quadratic equations class 10 questions require equation formation first. CBSE 2026 problems often use area, age, speed, and geometry conditions.

Q1. The Sum Of Ages Of Two Friends Is 20 Years. Four Years Ago, Their Age Product Was 48. Is This Possible?

The situation is not possible. The quadratic equation has no real roots.

1. Given Data:
   Present ages = x and 20 - x
   Product four years ago = 48
2. Equation Formation:
   (x - 4)(20 - x - 4) = 48
   (x - 4)(16 - x) = 48
3. Calculation:
   16x - x² - 64 + 4x = 48
   -x² + 20x - 112 = 0
   x² - 20x + 112 = 0
4. Discriminant:
   D = (-20)² - 4(1)(112)
   D = 400 - 448
   D = -48
5. Final Result:
   The situation is not possible

Q2. A Rectangular Mango Grove Has Length Twice Its Breadth And Area 800 m². Find Dimensions.

The breadth is 20 m, and the length is 40 m. The area condition gives a quadratic equation.

1. Given Data:
   Breadth = x m
   Length = 2x m
   Area = 800 m²
2. Formula Used:
   Area = length × breadth
3. Calculation:
   2x × x = 800
   2x² = 800
   x² = 400
   x = 20
4. Final Result:
   Breadth = 20 m, length = 40 m

Q3. A Rectangular Park Has Perimeter 80 m And Area 400 m². Find Dimensions.

The park is 20 m × 20 m. The discriminant equals zero, so both dimensions are equal.

1. Given Data:
   Perimeter = 80 m
   Area = 400 m²
2. Equation Formation:
   Let length = L and breadth = B.
   L + B = 40
   B = 40 - L
3. Calculation:
   L(40 - L) = 400
   40L - L² = 400
   L² - 40L + 400 = 0
4. Discriminant:
   D = 1600 - 1600
   D = 0
   L = 40/2 = 20
   B = 20
5. Final Result:
   Dimensions = 20 m × 20 m

Q4. A Pole Stands On The Boundary Of A Circular Park Of Diameter 13 m. Its Distances From Two Opposite Gates Differ By 7 m. Find Distances.

The distances are 5 m and 12 m. Use the right angle in a semicircle.

1. Given Data:
   Diameter AB = 13 m
   BP = x m
   AP = x + 7 m
2. Formula Used:
   AP² + BP² = AB²
3. Calculation:
   (x + 7)² + x² = 13²
   x² + 14x + 49 + x² = 169
   2x² + 14x - 120 = 0
   x² + 7x - 60 = 0
   (x - 5)(x + 12) = 0
4. Final Result:
   BP = 5 m and AP = 12 m

Q5. A Cottage Industry Produces Pottery Articles. Cost Per Article Is ₹3 More Than Twice The Number Produced. Total Cost Is ₹90. Find Both.

The industry produces 6 articles, and each costs ₹15. Form the total cost equation.

1. Given Data:
   Number of articles = x
   Cost per article = 2x + 3
   Total cost = ₹90
2. Equation Formation:
   x(2x + 3) = 90
3. Calculation:
   2x² + 3x - 90 = 0
   D = 3² - 4(2)(-90)
   D = 729
   x = [-3 + 27]/4
   x = 6
4. Cost:
   Cost per article = 2(6) + 3
   Cost per article = ₹15
5. Final Result:
   6 articles, ₹15 per article

Quadratic Equations Class 10 Extra Questions With Solutions

These quadratic equations class 10 extra questions cover value-based, discriminant-based, and word-problem patterns. Each question stays within the current NCERT 2026 scope.

Q1. If x = -1/2 Is A Solution Of 3x² + 2kx - 3 = 0, Find k.

The value of k is -9/4. Substitute x = -1/2 in the equation.

1. Given Data:
   3x² + 2kx - 3 = 0
   x = -1/2
2. Formula Used:
   Substitute the root in the equation.
3. Calculation:
   3(1/4) + 2k(-1/2) - 3 = 0
   3/4 - k - 3 = 0
   -k = 3 - 3/4
   -k = 9/4
4. Final Result:
   k = -9/4

Q2. If -5 Is A Root Of 2x² + px - 15 = 0 And p(x² + x) + k = 0 Has Equal Roots, Find k.

The value of k is 7/4. First find p, then use D = 0.

1. Given Data:
   -5 is a root of 2x² + px - 15 = 0
2. Find p:
   2(25) + p(-5) - 15 = 0
   50 - 5p - 15 = 0
   p = 7
3. New Equation:
   p(x² + x) + k = 0
   7x² + 7x + k = 0
4. Equal Roots Condition:
   D = 0
   7² - 4(7)(k) = 0
   49 - 28k = 0
5. Final Result:
   k = 7/4

Q3. Find k If x² + k(2x + k - 1) + 2 = 0 Has Real And Equal Roots.

The value of k is 2. Expand the equation first.

1. Given Data:
   x² + k(2x + k - 1) + 2 = 0
2. Standard Form:
   x² + 2kx + k² - k + 2 = 0
3. Formula Used:
   D = 0
4. Calculation:
   (2k)² - 4(1)(k² - k + 2) = 0
   4k² - 4k² + 4k - 8 = 0
   4k = 8
5. Final Result:
   k = 2

Q4. The Area Of A Rectangular Plot Is 528 m². Length Is One More Than Twice Its Breadth. Find Dimensions.

The breadth is 16 m, and the length is 33 m.

1. Given Data:
   Breadth = x m
   Length = 2x + 1 m
   Area = 528 m²
2. Equation Formation:
   x(2x + 1) = 528
3. Calculation:
   2x² + x - 528 = 0
   D = 1 + 4224
   D = 4225
   √4225 = 65
4. Find x:
   x = (-1 + 65)/4
   x = 16
   Length = 2(16) + 1
   Length = 33
5. Final Result:
   Breadth = 16 m, length = 33 m

Q5. Find The Discriminant Of 4x² - 5 = 0 And State The Nature Of Roots.

The discriminant is 80, so the roots are real and distinct.

1. Given Data:
   a = 4
   b = 0
   c = -5
2. Formula Used:
   D = b² - 4ac
3. Calculation:
   D = 0² - 4(4)(-5)
   D = 80
4. Final Result:
   Roots are real and distinct

Q6. Find p If One Root Of px² - 14x + 8 = 0 Is 6 Times The Other.

The value of p is 3. Use sum and product of roots.

1. Given Data:
   Roots = α and 6α
   Equation = px² - 14x + 8 = 0
2. Formula Used:
   Sum of roots = -b/a
   Product of roots = c/a
3. Sum:
   α + 6α = 7α
   7α = 14/p
   α = 2/p
4. Product:
   α × 6α = 8/p
   6α² = 8/p
   6(2/p)² = 8/p
   24/p² = 8/p
5. Final Result:
   p = 3

Class 10 Quadratic Equations Previous Year Questions With Solutions

These class 10 quadratic equations previous year questions follow repeated board-style patterns. They test equation formation, discriminant application, and root conditions.

Q1. A Train Travels 480 km At Uniform Speed. If Speed Were 8 km/h Less, It Would Take 3 Hours More. Find Speed.

The speed is 40 km/h. Form an equation using time = distance/speed.

1. Given Data:
   Distance = 480 km
   Original speed = x km/h
   Reduced speed = x - 8 km/h
   Extra time = 3 hours
2. Equation Formation:
   480/(x - 8) - 480/x = 3
3. Calculation:
   480[x - (x - 8)] / x(x - 8) = 3
   3840 / x(x - 8) = 3
   x(x - 8) = 1280
   x² - 8x - 1280 = 0
4. Solve:
   D = 64 + 5120 = 5184
   √5184 = 72
   x = (8 + 72)/2
5. Final Result:
   Speed = 40 km/h

Q2. Find The Discriminant Of 4x² - 5 = 0 And Comment On The Nature Of Roots.

The discriminant is 80, so the roots are real and distinct.

1. Given Data:
   a = 4
   b = 0
   c = -5
2. Formula Used:
   D = b² - 4ac
3. Calculation:
   D = 0² - 4(4)(-5)
   D = 80
4. Final Result:
   Two distinct real roots

Q3. Find p For Which px(x - 2) + 6 = 0 Has Two Equal Real Roots.

The value of p is 6. Reject p = 0 because the equation would not remain quadratic.

1. Given Data:
   px(x - 2) + 6 = 0
2. Standard Form:
   px² - 2px + 6 = 0
3. Formula Used:
   D = 0
4. Calculation:
   (-2p)² - 4(p)(6) = 0
   4p² - 24p = 0
   4p(p - 6) = 0
   p = 0 or p = 6
5. Final Result:
   p = 6

Q4. Find k So That x² - 4kx + k = 0 Has Equal Roots.

The values are k = 0 or k = 1/4. Equal roots occur when D = 0.

1. Given Data:
   x² - 4kx + k = 0
2. Formula Used:
   D = 0
3. Calculation:
   (-4k)² - 4(1)(k) = 0
   16k² - 4k = 0
   4k(4k - 1) = 0
4. Final Result:
   k = 0 or k = 1/4

Quadratic Equations Class 10 Questions And Answers: Most Repeated Variations

These quadratic equations class 10 questions and answers cover recurring CBSE 2026 patterns. Students should practise standard form, factorisation, discriminant, and word problems.

Q1. What Are The Most Repeated Variations In Quadratic Equations Class 10 Important Questions?

The most repeated variations are standard form, factorisation, nature of roots, finding k, area problems, and speed problems.

1. Factorisation Pattern:
   Split the middle term and find roots.
2. Discriminant Pattern:
   Use D = b² - 4ac.
3. Word Problem Pattern:
   Convert the condition into ax² + bx + c = 0.
4. Final Result:
   These six patterns cover most Chapter 4 board-style questions

Q2. How Should Students Solve Important Questions Class 10 Maths Chapter 4 With Solutions?

Students should first convert the question into standard form. The next step depends on factorisation, discriminant, or formula use.

1. Write Standard Form:
   Convert the equation to ax² + bx + c = 0.
2. Choose Method:
   Use factorisation when factors are easy.
3. Use Discriminant:
   Use D for root nature or equal-root conditions.
4. Final Result:
   Correct standard form decides the solution path

Useful Important Questions Class 10 Maths Links