Hello students how are you? hope you well ,,sorry for some delay in class 10 maths chapter 4 Quadratic equations important questions because i was suffring from fever. today i will provide you very important questions and their solutions step by step in my handwriting.
Preparing for the 2026 Board Exams requires more than just skimming through textbooks; it requires a strategic focus on high-yield topics. Chapter 4: Quadratic Equations is a vital part of the Class 10 Mathematics syllabus, often carrying a significant weightage of 5 to 9 marks. Whether it’s finding the value of ‘k’ or solving complex speed-distance word problems, this chapter tests your logical and algebraic skills.
In this guide, we provide a curated list of important questions with solutions, a comprehensive formula bank, and a downloadable PDF to ensure you are exam-ready.
Table of Contents
Quadratic Equations Class 10 Formula Bank & Key Concepts
you must first memorize the “Pillars of Quadratics
The Standard Form: Every quadratic equation must be written as where
The Quadratic Formula (Shridharacharya’s Rule): The roots of the equation are found using:
The Discriminant (D): This value, , tells you the nature of the roots before you even solve the equation.
Understanding the Nature of Roots (Most Repeated Questions)
The CBSE and state boards frequently ask 1-mark or 2-mark questions based on the Nature of Roots.
| Value of Discriminant (D) | Nature of Roots | Exam Relevance |
| D > 0 | Two Distinct Real Roots | Found in parabola-based questions. |
| D = 0 | Two Equal Real Roots | Most common for “Find the value of “k” questions. |
| D < 0 | No Real Roots | Often used to check if a situation is possible. |
What are Quadratic Equations?
A Quadratic Equation is a second-degree algebraic equation in a single variable where the highest power of the variable is two. In its Standard Form, it is represented as:
where a, b, and c are real numbers and the coefficient a is not equal to zero
Important questions
1.Find the values of k for which the quadratic equation has equal roots.
solution:
For an equation to have equal roots
Here, a = (k + 4), b = (k + 1), c = 1
Detailed Solution:
Given Equation: (k + 4)x² + (k + 1)x + 1 = 0
Comparing this with the standard quadratic form ax² + bx + c = 0, we identify the coefficients:
a = (k + 4), b = (k + 1), c = 1
For a quadratic equation to have real and equal roots, its Discriminant (D) must be exactly equal to zero:
D = b² – 4ac = 0
Substituting the values of a, b, and c:
(k + 1)² – 4(k + 4)(1) = 0
Expanding the expression using algebraic identity (A + B)² = A² + 2AB + B²:
(k² + 2k + 1) – (4k + 16) = 0
k² + 2k + 1 – 4k – 16 = 0
k² – 2k – 15 = 0
Now, we solve this quadratic equation in terms of k by splitting the middle term (-2k into -5k + 3k):
k² – 5k + 3k – 15 = 0
k(k – 5) + 3(k – 5) = 0
(k – 5)(k + 3) = 0
Setting each factor to zero:
k – 5 = 0 => k = 5
k + 3 = 0 => k = -3
Final Answer: The values of k for which the equation has equal roots are k = 5 or k = -3.
Q2. Solve the following quadratic equation for x by factorization: 4√3x² + 5x – 2√3 = 0
Detailed Solution:
Given Equation: 4√3x² + 5x – 2√3 = 0
To solve this by factorization (splitting the middle term), we find two numbers whose:
Sum = Coeff. of x = 5
Product = (Constant term) × (Coeff. of x²) = (-2√3) × (4√3) = -8 × 3 = -24
Two such numbers whose product is -24 and whose sum is 5 are 8 and -3.
Splitting the middle term 5x into 8x – 3x:
4√3x² + 8x – 3x – 2√3 = 0
Grouping terms to factor out common expressions:
4x(√3x + 2) – √3(√3x + 2) = 0 [Since 3 = √3 × √3]
Taking the common binomial factor (√3x + 2) out:
(√3x + 2)(4x – √3) = 0
Equating each factor to zero to find the roots:
Case 1: √3x + 2 = 0 => √3x = -2 => x = -2 / √3
Case 2: 4x – √3 = 0 => 4x = √3 => x = √3 / 4
Final Answer: The required roots of the equation are x = -2/√3 and x = √3/4.
Q3. A fast train takes 3 hours less than a slow train for a journey of 600 km. If the speed of the slow train is 10 km/h less than that of the fast train, find the uniform speeds of the two trains.
Detailed Solution:
Let the uniform speed of the fast train be x km/h.
Given that the speed of the slow train is 10 km/h less than the fast train:
Speed of the slow train = (x – 10) km/h
Total distance to cover = 600 km.
Using the basic formula Time = Distance / Speed:
Time taken by the fast train (T₁) = 600 / x hours
Time taken by the slow train (T₂) = 600 / (x – 10) hours
According to the problem, the difference in travel times is 3 hours:
T₂ – T₁ = 3
600 / (x – 10) – 600 / x = 3
Factoring out 600 and taking a common denominator:
600 * [ (x – (x – 10)) / (x(x – 10)) ] = 3
600 * [ 10 / (x² – 10x) ] = 3
6000 = 3(x² – 10x)
Dividing both sides by 3:
2000 = x² – 10x
x² – 10x – 2000 = 0
Solving the quadratic equation by splitting the middle term (-10x into -50x + 40x):
x² – 50x + 40x – 2000 = 0
x(x – 50) + 40(x – 50) = 0
(x – 50)(x + 40) = 0
This yields x = 50 or x = -40.
Since the speed of a train can never be negative, we reject x = -40. Thus, x = 50.
Speed of the fast train = 50 km/h
Speed of the slow train = (50 – 10) = 40 km/h
Final Answer: The speed of the fast train is 50 km/h and the speed of the slow train is 40 km/h.
Q4. Two water taps together can fill a tank in 9(3/8) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Detailed Solution:
Let the time taken by the smaller diameter tap to fill the tank completely be x hours.
The larger tap takes 10 hours less than the smaller tap:
Time taken by the larger diameter tap = (x – 10) hours
Therefore, the work rate per hour for each tap is:
Part of the tank filled by smaller tap in 1 hour = 1 / x
Part of the tank filled by larger tap in 1 hour = 1 / (x – 10)
Combined part filled by both taps in 1 hour = 1/x + 1/(x – 10)
The total time given for both taps working together is 9(3/8) hours = 75 / 8 hours.
Thus, the combined part filled in 1 hour is the reciprocal of total time = 8 / 75.
Setting up the equation:
1/x + 1/(x – 10) = 8 / 75
[ (x – 10 + x) / (x(x – 10)) ] = 8 / 75
(2x – 10) / (x² – 10x) = 8 / 75
Cross-multiplying to eliminate fractions:
75(2x – 10) = 8(x² – 10x)
150x – 750 = 8x² – 80x
Rearranging all terms to one side to form a standard quadratic equation:
8x² – 80x – 150x + 750 = 0
8x² – 230x + 750 = 0
Dividing the entire equation by 2 to simplify:
4x² – 115x + 375 = 0
Using the factor method, we find two numbers whose product is 4 × 375 = 1500 and whose sum is -115. These numbers are -100 and -15:
4x² – 100x – 15x + 375 = 0
4x(x – 25) – 15(x – 25) = 0
(4x – 15)(x – 25) = 0
This gives two potential cases:
Case 1: x – 25 = 0 => x = 25
Case 2: 4x – 15 = 0 => x = 15/4 = 3.75 hours
If x = 3.75 hours, then the time taken by the larger tap would be (3.75 – 10) = -6.25 hours. Since time taken cannot be negative, we reject x = 3.75.
Therefore, x = 25 hours.
Time taken by smaller tap = 25 hours
Time taken by larger tap = (25 – 10) = 15 hours
Final Answer: The smaller tap takes 25 hours and the larger tap takes 15 hours to separately fill the tank.
Q5. The hypotenuse of a right-angled triangle is 6 m more than twice the shortest side. If the third side is 2 m less than the hypotenuse, find the exact lengths of all three sides of the triangle.
Detailed Solution:
Let the length of the shortest side of the right-angled triangle be x meters.
According to the given conditions:
Hypotenuse = 2x + 6
Third side = (Hypotenuse) – 2 = (2x + 6) – 2 = 2x + 4
By applying Pythagoras Theorem (Base² + Perpendicular² = Hypotenuse²):
x² + (2x + 4)² = (2x + 6)²
Expanding the binomial terms using identity (A + B)² = A² + 2AB + B²:
x² + (4x² + 16x + 16) = 4x² + 24x + 36
5x² + 16x + 16 = 4x² + 24x + 36
Shifting all terms to the left-hand side:
5x² – 4x² + 16x – 24x + 16 – 36 = 0
x² – 8x – 20 = 0
Solving the equation by splitting the middle term (-8x into -10x + 2x):
x² – 10x + 2x – 20 = 0
x(x – 10) + 2(x – 10) = 0
(x – 10)(x + 2) = 0
This yields two roots: x = 10 or x = -2.
Since a side length can never be negative, we discard x = -2. Therefore, x = 10.
Calculating the lengths of all sides:
Shortest side = x = 10 m
Hypotenuse = 2(10) + 6 = 20 + 6 = 26 m
Third side = 2(10) + 4 = 20 + 4 = 24 m
Final Answer: The lengths of the sides of the triangle are 10 m, 24 m, and 26 m.
Q6. Determine the range of values of p for which the quadratic equation px² – 6x + (p – 8) = 0 possesses two distinct and real roots.
Complete Analytical Solution:
Given Equation: px² – 6x + (p – 8) = 0
First, we match the coefficients with the standard form Ax² + Bx + C = 0:
A = p, B = -6, C = (p – 8)
Crucial Constraint: For this to remain a valid quadratic equation, the coefficient of x² cannot be zero. Therefore, p ≠ 0.
Condition for Roots: For a quadratic equation to have distinct and real roots, its Discriminant (D) must be strictly greater than zero:
D = B² – 4AC > 0
Substituting our coefficients into the inequality:
(-6)² – 4(p)(p – 8) > 0
36 – 4p(p – 8) > 0
Expanding the expression:
36 – 4p² + 32p > 0
Rearranging terms to put it in standard polynomial form:
-4p² + 32p + 36 > 0
Dividing the entire inequality by -4. Remember, dividing an inequality by a negative number reverses the direction of the inequality sign:
p² – 8p – 9 < 0
Factoring the quadratic expression by splitting the middle term (-8p into -9p + p):
p² – 9p + p – 9 < 0
p(p – 9) + 1(p – 9) < 0
(p – 9)(p + 1) < 0
For the product of two linear terms to be negative, the variable must lie strictly between the two critical roots (-1 and 9):
-1 < p < 9
Recalling our critical initial constraint that p ≠ 0:
Final Answer: The range of values for p is -1 < p < 9 excluding p = 0 (or written as: p ∈ (-1, 9) except {0}).
Q7. A local school group purchased a batch of uniform study books for a total cost of ₹1,800. If each book had cost ₹10 less, the group could have acquired 6 more books for the exact same budget. Find the original count of books purchased.
Complete Analytical Solution:
Let the original number of study books purchased be n.
Total capital budget assigned = ₹1,800.
Therefore, the baseline price evaluated per individual unit book is:
Original Cost Per Book = 1800 / n
According to the alternative hypothetical scenario, if individual pricing falls by ₹10:
New Price Per Book = (1800 / n) – 10
With this modified cost per book, the absolute quantity of books attainable increases by 6:
New Total Book Count = n + 6
Since the absolute financial allocation pool remains locked at ₹1,800, our balanced equation is:
New Total Book Count × New Price Per Book = Total Budget
(n + 6) * [ (1800 / n) – 10 ] = 1800
Expanding the binomial products explicitly:
n*(1800 / n) – 10n + 6*(1800 / n) – 60 = 1800
1800 – 10n + 10800 / n – 60 = 1800
Subtracting 1800 from both balancing horizons:
-10n + 10800 / n – 60 = 0
Multiply the complete baseline calculation structure by n to completely clean out fractions:
-10n² + 10800 – 60n = 0
Rearranging terms and multiplying by -1 to normalize the leading variable coefficient sign:
10n² + 60n – 10800 = 0
Divide the entire sequence by 10 to clear excess constants:
n² + 6n – 1080 = 0
Factor the equation via split terms. We seek factor parameters matching a product of -1080 and delta variance of +6. The numbers are +36 and -30:
n² + 36n – 30n – 1080 = 0
n(n + 36) – 30(n + 36) = 0
(n + 36)(n – 30) = 0
This establishes our roots: n = 30 or n = -36.
Physical constraint rule: Since counts of physical books purchased cannot logically fall below zero, we discard n = -36 completely.
Final Answer: The original number of books purchased by the school group is 30.
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Conclusion
Mastering Chapter 4: Quadratic Equations is one of the most effective ways to secure high marks in your CBSE Class 10 Mathematics board exam. From understanding the core structure (ax2 + bx + c = 0) to mastering the nature of roots using the discriminant formula (D = b2– 4ac), this chapter forms the foundation for higher-level algebra.