Class 10 Maths Chapter 4 Exercise 4.2 Solutions English Medium | Quadratic Equations SEBA 2026–2027
Exercise 4.2 is one of the most important parts of the chapter Quadratic Equations. In this exercise, students learn how to solve quadratic equations using the Quadratic Formula, which is extremely useful when factorization becomes difficult or impossible.
At Digital Pipal Academy, our experienced teachers have explained every step in a clear and exam-oriented way so that students can easily understand and score full marks.
💡 Important Tips for Students
✔ Always write the equation in standard form
✔ Identify a, b, c correctly
✔ Calculate D carefully
✔ Don’t forget “±” sign
✔ Simplify your final answer
🎯 Exam Importance
📌 Very important for Board Exams
📌 3–4 mark questions are commonly asked
📌 Step-by-step method ensures full marks
🌐 Complete Exercise 4.2 Solutions
(i) \(x^2 - 3x - 10 = 0\)
(ii) \(2x^2 + x - 6 = 0\)
(iii) \(\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0\)
(iv) \(2x^2 - x + \frac{1}{8} = 0\)
(v) \(100x^2 - 20x + 1 = 0\)
(vi) \(2x^2 - 7x + 6 = 0\)
(vii) \(x^2 - 10x - 96 = 0\)
(viii) \(\sqrt{3}x^2 + 10x + 7\sqrt{3} = 0\)
(ix) \(x^2 + 2\sqrt{2}x + 2 = 0\)
(x) \(14x + 5 - 3x^2 = 0\)
Solution:
(i) \(x^2 - 3x - 10 = 0\)
\(\Rightarrow x^2 - 5x + 2x - 10 = 0\)
\(\Rightarrow x(x - 5) + 2(x - 5) = 0\)
\(\Rightarrow (x - 5)(x + 2) = 0\)
Either \(x - 5 = 0 \Rightarrow x = 5\) or \(x + 2 = 0 \Rightarrow x = -2\).
Roots: \(5, -2\)
(ii) \(2x^2 + x - 6 = 0\)
\(\Rightarrow 2x^2 + 4x - 3x - 6 = 0\)
\(\Rightarrow 2x(x + 2) - 3(x + 2) = 0\)
\(\Rightarrow (x + 2)(2x - 3) = 0\)
Either \(x + 2 = 0 \Rightarrow x = -2\) or \(2x - 3 = 0 \Rightarrow x = \frac{3}{2}\).
Roots: \(-2, \frac{3}{2}\)
(iii) \(\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0\)
\(\Rightarrow \sqrt{2}x^2 + 5x + 2x + 5\sqrt{2} = 0\)
\(\Rightarrow x(\sqrt{2}x + 5) + \sqrt{2}(\sqrt{2}x + 5) = 0\)
\(\Rightarrow (\sqrt{2}x + 5)(x + \sqrt{2}) = 0\)
Either \(x = -\frac{5}{\sqrt{2}}\) or \(x = -\sqrt{2}\).
Roots: \(-\frac{5}{\sqrt{2}}, -\sqrt{2}\)
(iv) \(2x^2 - x + \frac{1}{8} = 0\)
Multiply by 8: \(16x^2 - 8x + 1 = 0\)
\(\Rightarrow 16x^2 - 4x - 4x + 1 = 0\)
\(\Rightarrow 4x(4x - 1) - 1(4x - 1) = 0\)
\(\Rightarrow (4x - 1)(4x - 1) = 0\)
Roots: \(\frac{1}{4}, \frac{1}{4}\)
(v) \(100x^2 - 20x + 1 = 0\)
\(\Rightarrow 100x^2 - 10x - 10x + 1 = 0\)
\(\Rightarrow 10x(10x - 1) - 1(10x - 1) = 0\)
\(\Rightarrow (10x - 1)^2 = 0\)
Roots: \(\frac{1}{10}, \frac{1}{10}\)
(vi) \(2x^2 - 7x + 6 = 0\)
\(\Rightarrow 2x^2 - 4x - 3x + 6 = 0\)
\(\Rightarrow 2x(x - 2) - 3(x - 2) = 0\)
\(\Rightarrow (x - 2)(2x - 3) = 0\)
Roots: \(2, \frac{3}{2}\)
(vii) \(x^2 - 10x - 96 = 0\)
\(\Rightarrow x^2 - 16x + 6x - 96 = 0\)
\(\Rightarrow x(x - 16) + 6(x - 16) = 0\)
\(\Rightarrow (x - 16)(x + 6) = 0\)
Roots: \(16, -6\)
(viii) \(\sqrt{3}x^2 + 10x + 7\sqrt{3} = 0\)
\(\Rightarrow \sqrt{3}x^2 + 3x + 7x + 7\sqrt{3} = 0\)
\(\Rightarrow \sqrt{3}x(x + \sqrt{3}) + 7(x + \sqrt{3}) = 0\)
\(\Rightarrow (x + \sqrt{3})(\sqrt{3}x + 7) = 0\)
Roots: \(-\sqrt{3}, -\frac{7}{\sqrt{3}}\)
(ix) \(x^2 + 2\sqrt{2}x + 2 = 0\)
\(\Rightarrow x^2 + \sqrt{2}x + \sqrt{2}x + 2 = 0\)
\(\Rightarrow x(x + \sqrt{2}) + \sqrt{2}(x + \sqrt{2}) = 0\)
\(\Rightarrow (x + \sqrt{2})^2 = 0\)
Roots: \(-\sqrt{2}, -\sqrt{2}\)
(x) \(14x + 5 - 3x^2 = 0\)
Rearrange: \(-3x^2 + 14x + 5 = 0 \Rightarrow 3x^2 - 14x - 5 = 0\)
\(\Rightarrow 3x^2 - 15x + x - 5 = 0\)
\(\Rightarrow 3x(x - 5) + 1(x - 5) = 0\)
\(\Rightarrow (x - 5)(3x + 1) = 0\)
Roots: \(5, -\frac{1}{3}\)
Represent the following situations mathematically:
(i) John and Jayanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs 750. We would like to find out the number of toys produced on that day.
Solution:
(i)
Let the number of marbles John had = \(x\).Since the total number of marbles is 45, the number of marbles Jivanti had = \(45 - x\).
After both lost 5 marbles each:
Number of marbles John has = \(x - 5\)
Number of marbles Jivanti has = \(45 - x - 5 = 40 - x\)
According to the problem, the product of the number of marbles they now have is 124:
\((x - 5)(40 - x) = 124\)
\(\Rightarrow 40x - x^2 - 200 + 5x = 124\)
\(\Rightarrow -x^2 + 45x - 200 - 124 = 0\)
\(\Rightarrow -x^2 + 45x - 324 = 0\)
\(\Rightarrow x^2 - 45x + 324 = 0\)
Now, by splitting the middle term:
\(\Rightarrow x^2 - 36x - 9x + 324 = 0\)
\(\Rightarrow x(x - 36) - 9(x - 36) = 0\)
\(\Rightarrow (x - 36)(x - 9) = 0\)
So, \(x = 36\) or \(x = 9\).
Therefore, they initially had 36 and 9 marbles respectively.
(ii)
Let the number of toys produced on that day = \(x\).The cost of production of each toy = \((55 - x)\) Rupees.
According to the problem, the total cost of production on that day was 750:
\(x(55 - x) = 750\)
\(\Rightarrow 55x - x^2 = 750\)
\(\Rightarrow -x^2 + 55x - 750 = 0\)
\(\Rightarrow x^2 - 55x + 750 = 0\)
Now, by splitting the middle term:
\(\Rightarrow x^2 - 30x - 25x + 750 = 0\)
\(\Rightarrow x(x - 30) - 25(x - 30) = 0\)
\(\Rightarrow (x - 30)(x - 25) = 0\)
So, \(x = 30\) or \(x = 25\).
Therefore, the number of toys produced on that day was either 30 or 25.
Solution:
Let the first number be \(x\).
Since the sum of the two numbers is 27, the second number will be \((27 - x)\).
According to the problem, the product of these two numbers is 182. Therefore:
\(x(27 - x) = 182\)
\(\Rightarrow 27x - x^2 = 182\)
\(\Rightarrow -x^2 + 27x - 182 = 0\)
\(\Rightarrow x^2 - 27x + 182 = 0\)
Now, we split the middle term (finding two numbers that multiply to 182 and add up to 27, which are 13 and 14):
\(\Rightarrow x^2 - 14x - 13x + 182 = 0\)
\(\Rightarrow x(x - 14) - 13(x - 14) = 0\)
\(\Rightarrow (x - 14)(x - 13) = 0\)
So,
Either \(x - 14 = 0 \Rightarrow x = 14\)
Or \(x - 13 = 0 \Rightarrow x = 13\)
If the first number is 14, the second number is \(27 - 14 = 13\). If the first number is 13, the second number is \(27 - 13 = 14\).
Therefore, the two required numbers are 13 and 14.
Solution:
Let the first positive integer be \(x\).
The next consecutive positive integer will be \((x + 1)\).
According to the question, the sum of their squares is 365:
\(x^2 + (x + 1)^2 = 365\)
\(\Rightarrow x^2 + (x^2 + 2x + 1) = 365\)
\(\Rightarrow 2x^2 + 2x + 1 - 365 = 0\)
\(\Rightarrow 2x^2 + 2x - 364 = 0\)
Dividing by 2:
\(\Rightarrow x^2 + x - 182 = 0\)
Splitting the middle term (since \(14 \times 13 = 182\)):
\(\Rightarrow x^2 + 14x - 13x - 182 = 0\)
\(\Rightarrow x(x + 14) - 13(x + 14) = 0\)
\(\Rightarrow (x + 14)(x - 13) = 0\)
Either \(x + 14 = 0 \Rightarrow x = -14\) (Rejected, as we need a positive integer)
Or \(x - 13 = 0 \Rightarrow x = 13\).
Therefore, the two consecutive positive integers are 13 and 14.
Solution:
Let the base of the right triangle be \(x\) cm.
Then, the altitude (height) will be \((x - 7)\) cm.
Hypotenuse = 13 cm.
Using Pythagoras Theorem (\(Base^2 + Altitude^2 = Hypotenuse^2\)):
\(x^2 + (x - 7)^2 = 13^2\)
\(\Rightarrow x^2 + (x^2 - 14x + 49) = 169\)
\(\Rightarrow 2x^2 - 14x + 49 - 169 = 0\)
\(\Rightarrow 2x^2 - 14x - 120 = 0\)
Dividing by 2:
\(\Rightarrow x^2 - 7x - 60 = 0\)
Splitting the middle term (since \(12 \times 5 = 60\)):
\(\Rightarrow x^2 - 12x + 5x - 60 = 0\)
\(\Rightarrow x(x - 12) + 5(x - 12) = 0\)
\(\Rightarrow (x - 12)(x + 5) = 0\)
Either \(x - 12 = 0 \Rightarrow x = 12\)
Or \(x + 5 = 0 \Rightarrow x = -5\) (Side cannot be negative).
Base = 12 cm and Altitude = \(12 - 7 = 5\) cm.
Solution:
Let the number of articles produced be \(x\).
Cost of each article = \((2x + 3)\) Rupees.
Total cost = Number of articles \(\times\) Cost of each article = 90:
\(x(2x + 3) = 90\)
\(\Rightarrow 2x^2 + 3x - 90 = 0\)
Splitting the middle term (since \(15 \times 12 = 180\)):
\(\Rightarrow 2x^2 + 15x - 12x - 90 = 0\)
\(\Rightarrow x(2x + 15) - 6(2x + 15) = 0\)
\(\Rightarrow (2x + 15)(x - 6) = 0\)
Either \(2x + 15 = 0 \Rightarrow x = -7.5\) (Rejected)
Or \(x - 6 = 0 \Rightarrow x = 6\).
Number of articles = 6
Cost of each article = \(2(6) + 3 = 15\) Rupees.
(b) \(-1\)
(c) \(1\)
(d) \(2\)
Solution:
The given equation is \(x^2 - 2px + p^2 = 0\).
This is in the form of the algebraic identity \((a - b)^2 = a^2 - 2ab + b^2\).
\(\Rightarrow (x - p)^2 = 0\)
Taking square root on both sides:
\(\Rightarrow x - p = 0\)
\(\Rightarrow x = p\)
Now, we need to find the value of \(\frac{p}{x}\):
\(\Rightarrow \frac{p}{p} = 1\)
Correct Option: (c) \(1\)
Step 1 : \(x^2 - 3x - 10 = 0\)
Step 2 : \(x^2 - 5x - 2x - 10 = 0\)
Step 3 : \(x(x - 5) - 2(x - 5) = 0\)
Step 4 : \((x - 5)(x - 2) = 0\)
Step 5 : \(x = 5\) আৰু \(x = 2\)
Which step did the student make the first mistake
(b) Step : 3
(c) Step : 4
(d) Step : 5
Solution:
Let's look at Step 2 provided in the question:
Step 2: \(x^2 - 5x - 2x - 10 = 0\)
To split the middle term \(-3x\), we need two numbers whose sum is \(-3\) and product is \(-10\). Those numbers are \(-5\) and \(+2\).
Correct split: \(-5x + 2x = -3x\).
The student wrote \(-5x - 2x\), which equals \(-7x\).
Correct Option: (a) Step : 2
Solution:
Comparing \(2x^2 - 9x + 4 = 0\) with \(ax^2 + bx + c = 0\), we get:
\(a = 2, b = -9, c = 4\)
Sum of roots \((\alpha + \beta) = \frac{-b}{a} = \frac{-(-9)}{2} = \frac{9}{2}\)
Product of roots \((\alpha\beta) = \frac{c}{a} = \frac{4}{2} = 2\)
Solution:
Since \(x = 2\) is a root, it must satisfy the equation:
\(2(2)^2 + k(2) - 6 = 0\)
\(\Rightarrow 2(4) + 2k - 6 = 0\)
\(\Rightarrow 8 + 2k - 6 = 0\)
\(\Rightarrow 2k + 2 = 0\)
\(\Rightarrow 2k = -2\)
\(\Rightarrow k = -1\)
Now, the equation becomes \(2x^2 - x - 6 = 0\).
To find the other root, we know the product of roots \(= \frac{c}{a}\):
\(2 \times (\text{Other Root}) = \frac{-6}{2}\)
\(2 \times (\text{Other Root}) = -3\)
\(\Rightarrow \text{Other Root} = -\frac{3}{2}\)
Value of \(k = -1\) and the other root is \(-\frac{3}{2}\).
The total area of the lawn and the walking path is \(360\) square meters. The width of the walking path around the lawn is uniform. The dimensions of the lawn are \(12\) meters × \(10\) meters.
Based on the above information, answer the following questions:
(i) Taking the width of the walking path as \(x\) meters, form a quadratic equation representing the total area of the lawn and the walking path.
(ii) Solve the quadratic equation to find the width \(x\) of the walking path.
(iii) If the total cost of paving the walking path at the rate of ₹50 per square meter is ₹12,000, then find the area of the walking path.
(iv) Find the perimeter of the lawn.
Solution:
(i) Let the width of the walking path be \(x\) meters.
The dimensions of the inner lawn are \(12\) m × \(10\) m.
The outer dimensions (Lawn + Path) will be:
Outer Length = \((12 + 2x)\) m
Outer Breadth = \((10 + 2x)\) m
Given, total area = \(360\) m².
\((12 + 2x)(10 + 2x) = 360\)
\(\Rightarrow 120 + 24x + 20x + 4x^2 = 360\)
\(\Rightarrow 4x^2 + 44x - 240 = 0\)
Dividing by 4:
\(\Rightarrow x^2 + 11x - 60 = 0\)
(ii) From the equation \(x^2 + 11x - 60 = 0\):
\(\Rightarrow x^2 + 15x - 4x - 60 = 0\)
\(\Rightarrow x(x + 15) - 4(x + 15) = 0\)
\(\Rightarrow (x + 15)(x - 4) = 0\)
Either \(x + 15 = 0 \Rightarrow x = -15\) (Not possible)
Or \(x - 4 = 0 \Rightarrow x = 4\).
Width of the path, \(x = 4\) meters.
(iii) Total paving cost = ₹12,000
Rate = ₹50 per m²
Area of the path = \(\frac{\text{Total Cost}}{\text{Rate}}\)
\(\Rightarrow \text{Area} = \frac{12,000}{50}\)
Area of the path = \(240\) m².
(iv) Perimeter of the lawn = \(2 \times (\text{Length} + \text{Breadth})\)
\(\Rightarrow \text{Perimeter} = 2 \times (12 + 10)\)
\(\Rightarrow \text{Perimeter} = 2 \times 22\)
Perimeter of the lawn = \(44\) meters.
👉 Get all questions solved in detail on our website:
Visit: www.pipalacademy.com
❤️ Final Words
The Quadratic Formula is your most reliable method. Even when other methods fail, this formula will always help you solve the equation.
📘 Practice regularly and master the concept for guaranteed success.
Class 10 Maths Chapter 4 – Quadratic Equations (Exercise 4.2) FAQs
Exercise 4.2 focuses on solving quadratic equations using the factorization method.
It involves expressing the quadratic equation as a product of two linear factors and solving them.
This exercise is very important because factorization questions frequently appear in exams.
Digital Pipal Academy offers step-by-step solutions, concept clarity, updated syllabus content, and exam-focused guidance.
You can find complete solutions on Digital Pipal Academy in an easy and student-friendly format.
.png)
