📘 Class 10 Mathematics (English Medium)
Chapter 4: Quadratic Equations
Welcome to the complete solutions of Exercise 4.4 – Quadratic Equations for Class 10 Mathematics (English Medium).
Exercise 4.4 is one of the most important exercises of this chapter because it focuses on the nature of roots of quadratic equations using the Discriminant. Students learn how to determine whether the roots are real, equal, distinct, or imaginary without actually solving the equation completely.
At Digital Pipal Academy, our experienced teachers have prepared easy, detailed, and exam-oriented explanations to help students understand every concept clearly.
Get complete Class 10 Maths Chapter 4 Exercise 4.4 solutions in English Medium based on SEBA/SCERT Assam 2026–2027 syllabus. Learn step-by-step solutions of Quadratic Equations and word problems easily.
✨ Exercise 4.4 – Complete Step-by-Step Solutions
Conditions for Nature of Roots:
1. If \(b^2 - 4ac > 0\): Two distinct real roots.
2. If \(b^2 - 4ac = 0\): Two equal real roots.
3. If \(b^2 - 4ac < 0\): No real roots.
Solution:
(i) \(2x^2 - 3x + 5 = 0\)
Here, \(a = 2, b = -3, c = 5\).
Discriminant \(D = b^2 - 4ac = (-3)^2 - 4(2)(5) = 9 - 40 = -31\).
Since \(D < 0\), the equation has no real roots.
(ii) \(3x^2 - 4\sqrt{3}x + 4 = 0\)
Here, \(a = 3, b = -4\sqrt{3}, c = 4\).
Discriminant \(D = (-4\sqrt{3})^2 - 4(3)(4) = 48 - 48 = 0\).
Since \(D = 0\), the roots are real and equal.
Roots: \(x = \frac{-b}{2a} = \frac{-(-4\sqrt{3})}{2(3)} = \frac{4\sqrt{3}}{6} = \frac{2\sqrt{3}}{3}\).
Roots: \(\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}\)
(iii) \(2x^2 - 6x + 3 = 0\)
Here, \(a = 2, b = -6, c = 3\).
Discriminant \(D = (-6)^2 - 4(2)(3) = 36 - 24 = 12\).
Since \(D > 0\), the roots are real and distinct.
Using formula: \(x = \frac{-(-6) \pm \sqrt{12}}{2(2)} = \frac{6 \pm 2\sqrt{3}}{4} = \frac{3 \pm \sqrt{3}}{2}\).
Roots: \(\frac{3 + \sqrt{3}}{2}, \frac{3 - \sqrt{3}}{2}\)
(iv) \(9x^2 - 6x + 1 = 0\)
Here, \(a = 9, b = -6, c = 1\).
Discriminant \(D = (-6)^2 - 4(9)(1) = 36 - 36 = 0\).
Since \(D = 0\), the roots are real and equal.
\(x = \frac{-(-6)}{2(9)} = \frac{6}{18} = \frac{1}{3}\).
Roots: \(\frac{1}{3}, \frac{1}{3}\)
(v) \(3x^2 - 5x + 12 = 0\)
Here, \(a = 3, b = -5, c = 12\).
Discriminant \(D = (-5)^2 - 4(3)(12) = 25 - 144 = -119\).
Since \(D < 0\), there are no real roots.
(vi) \(x^2 + x + 1 = 0\)
Here, \(a = 1, b = 1, c = 1\).
Discriminant \(D = 1^2 - 4(1)(1) = -3\).
Since \(D < 0\), there are no real roots.
(vii) \(x^2 - 2\sqrt{3}x - 9 = 0\)
Here, \(a = 1, b = -2\sqrt{3}, c = -9\).
Discriminant \(D = (-2\sqrt{3})^2 - 4(1)(-9) = 12 + 36 = 48\).
Since \(D > 0\), roots are real and distinct.
\(x = \frac{2\sqrt{3} \pm \sqrt{48}}{2} = \frac{2\sqrt{3} \pm 4\sqrt{3}}{2}\).
Roots: \(3\sqrt{3}, -\sqrt{3}\)
(viii) \(2x^2 - 3\sqrt{3}x + 3 = 0\)
Here, \(a = 2, b = -3\sqrt{3}, c = 3\).
Discriminant \(D = (-3\sqrt{3})^2 - 4(2)(3) = 27 - 24 = 3\).
Since \(D > 0\), roots are real and distinct.
\(x = \frac{3\sqrt{3} \pm \sqrt{3}}{4}\).
Roots: \(\sqrt{3}, \frac{\sqrt{3}}{2}\)
Condition: A quadratic equation has two equal real roots if its discriminant \(D = b^2 - 4ac = 0\).
Solution:
(i) \(2x^2 + kx + 3 = 0\)
Here, \(a = 2, b = k, c = 3\).
For equal roots, \(b^2 - 4ac = 0\)
\(\Rightarrow k^2 - 4(2)(3) = 0\)
\(\Rightarrow k^2 - 24 = 0\)
\(\Rightarrow k^2 = 24\)
\(\Rightarrow k = \pm \sqrt{24} = \pm 2\sqrt{6}\)
Value of \(k\): \(\pm 2\sqrt{6}\)
(ii) \(kx(x - 2) + 6 = 0\)
Standard form: \(kx^2 - 2kx + 6 = 0\)
Here, \(a = k, b = -2k, c = 6\).
For equal roots, \(b^2 - 4ac = 0\)
\(\Rightarrow (-2k)^2 - 4(k)(6) = 0\)
\(\Rightarrow 4k^2 - 24k = 0\)
\(\Rightarrow 4k(k - 6) = 0\)
Since \(k\) cannot be 0 (as the equation must be quadratic), \(k - 6 = 0 \Rightarrow k = 6\).
Value of \(k\): 6
(iii) \(x^2 - (k + 4)x + 2k + 5 = 0\)
Here, \(a = 1, b = -(k + 4), c = 2k + 5\).
For equal roots, \(b^2 - 4ac = 0\)
\(\Rightarrow [-(k + 4)]^2 - 4(1)(2k + 5) = 0\)
\(\Rightarrow k^2 + 8k + 16 - 8k - 20 = 0\)
\(\Rightarrow k^2 - 4 = 0\)
\(\Rightarrow k^2 = 4 \Rightarrow k = \pm 2\)
Value of \(k\): \(\pm 2\)
(iv) \(2x^2 + 8x - k^3 = 0\)
Here, \(a = 2, b = 8, c = -k^3\).
For equal roots, \(b^2 - 4ac = 0\)
\(\Rightarrow 8^2 - 4(2)(-k^3) = 0\)
\(\Rightarrow 64 + 8k^3 = 0\)
\(\Rightarrow 8k^3 = -64\)
\(\Rightarrow k^3 = -8 \Rightarrow k = -2\)
Value of \(k\): -2
(v) \((k - 2)x^2 + 6x + 9 = 0\)
Here, \(a = k - 2, b = 6, c = 9\).
For equal roots, \(b^2 - 4ac = 0\)
\(\Rightarrow 6^2 - 4(k - 2)(9) = 0\)
\(\Rightarrow 36 - 36(k - 2) = 0\)
\(\Rightarrow 1 - (k - 2) = 0\)
\(\Rightarrow 1 - k + 2 = 0 \Rightarrow 3 - k = 0 \Rightarrow k = 3\).
Value of \(k\): 3
(vi) \((k - 12)x^2 + 2(k - 12)x + 2 = 0\)
Here, \(a = k - 12, b = 2(k - 12), c = 2\).
For equal roots, \(b^2 - 4ac = 0\)
\(\Rightarrow [2(k - 12)]^2 - 4(k - 12)(2) = 0\)
\(\Rightarrow 4(k - 12)^2 - 8(k - 12) = 0\)
\(\Rightarrow 4(k - 12) [ (k - 12) - 2 ] = 0\)
\(\Rightarrow 4(k - 12) (k - 14) = 0\)
Since \(k \ne 12\), we have \(k - 14 = 0 \Rightarrow k = 14\).
Value of \(k\): 14
Q3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is \(800 \text{ m}^2\)? If so, find its length and breadth.
Solution:
Let the breadth of the rectangular grove be \(x\) metres.Therefore, the length of the grove = \(2x\) metres.
According to the question, Area = \(800\)
\(\Rightarrow \text{Length} \times \text{Breadth} = 800\)
\(\Rightarrow 2x \times x = 800\)
\(\Rightarrow 2x^2 = 800\)
\(\Rightarrow x^2 = 400\)
\(\Rightarrow x = \sqrt{400} = 20\)
Since the value of \(x\) is a real number, it is possible to design such a grove.
Answer: The breadth is 20 m and the length is (2 \(\times\) 20) = 40 m.
Q4. Is the following situation possible? If so, determine their present ages.
The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Solution:
Let the present age of one friend be \(x\) years.Then, the present age of the other friend = \((20 - x)\) years.
Four years ago:
Age of the first friend = \((x - 4)\) years.
Age of the second friend = \((20 - x - 4) = (16 - x)\) years.
According to the question,
\((x - 4)(16 - x) = 48\)
\(\Rightarrow 16x - x^2 - 64 + 4x = 48\)
\(\Rightarrow -x^2 + 20x - 64 - 48 = 0\)
\(\Rightarrow x^2 - 20x + 112 = 0\)
Here, the Discriminant \(D = b^2 - 4ac\)
\(\Rightarrow D = (-20)^2 - 4(1)(112)\)
\(\Rightarrow D = 400 - 448 = -48\)
Since \(D < 0\), the equation has no real roots. Therefore, the given situation is not possible.
Q5. Is it possible to design a rectangular park of perimeter 80 m and area \(400 \text{ m}^2\)? If so, find its length and breadth.
Solution:
Let the length of the park be \(l\) and the breadth be \(b\).Perimeter = \(2(l + b) = 80 \Rightarrow l + b = 40\)
\(\therefore b = (40 - l)\)
Area = \(l \times b = 400\)
\(\Rightarrow l(40 - l) = 400\)
\(\Rightarrow 40l - l^2 = 400\)
\(\Rightarrow l^2 - 40l + 400 = 0\)
\(\Rightarrow (l - 20)^2 = 0\)
\(\Rightarrow l = 20\)
Since the value of \(l\) is a real number, it is possible to design the park.
Length = 20 m.
Breadth = (40 - 20) = 20 m.
Answer: The length and breadth are both 20 m (making it a square).
(b) There are two equal roots
(c) There are two distinct real roots
(d) There are more than two real roots
Solution:
Given, \(9x^2 - 6x - 2 = 0\)
Here, \(a = 9,\; b = -6,\; c = -2\)
Discriminant \(D = b^2 - 4ac\)
\(\Rightarrow D = (-6)^2 - 4(9)(-2)\)
\(\Rightarrow D = 36 + 72 = 108\)
Since \(D > 0\), the equation has two real and distinct roots.
Correct Answer: (c)
| Column I (Equation) | Column II (Nature of Roots) |
|---|---|
| P. \(x^2 - 2x + 3 = 0\) | (i) Roots are real and irrational |
| Q. \(3x^2 - 2x + \frac{1}{3} = 0\) | (ii) Roots are real and unequal |
| R. \(x^2 - 4x + 3 = 0\) | (iii) Roots are not real |
| S. \(2x^2 - x - 4 = 0\) | (iv) Roots are real and equal |
Analysis:
2. Q: \(D = (-2)^2 - 4(3)\left(\frac{1}{3}\right) = 4 - 4 = 0\) → (iv)
3. R: \(D = (-4)^2 - 4(1)(3) = 16 - 12 = 4 > 0\) → (ii)
4. S: \(D = (-1)^2 - 4(2)(-4) = 1 + 32 = 33 > 0\) → (i)
Choose the correct option:
(b) P → (iii), Q → (iv), R → (ii), S → (i)
(c) P → (iv), Q → (iii), R → (i), S → (ii)
(d) P → (ii), Q → (iv), R → (i), S → (iii)
Assertion (A): In the quadratic equation \((k - 1)x^2 - 10x + 3 = 0\), if one root is the reciprocal of the other, then the value of \(k\) is 4.
Reason (R): The roots of a quadratic equation \(ax^2 + bx + c = 0,\; (a \ne 0)\) are reciprocals of each other if \(a = c\).
Choose the correct option:
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
Solution and Analysis:
Verifying Reason (R):
Let the roots of the equation \(ax^2 + bx + c = 0\) be \(\alpha\) and \(\frac{1}{\alpha}\).
We know that, Product of roots = \(\frac{c}{a}\)
\(\Rightarrow \alpha \times \frac{1}{\alpha} = \frac{c}{a}\)
\(\Rightarrow 1 = \frac{c}{a}\)
\(\Rightarrow a = c\)
Therefore, Reason (R) is true.
Verifying Assertion (A):
Given equation: \((k - 1)x^2 - 10x + 3 = 0\)
Here, \(a = (k - 1)\) and \(c = 3\).
Since one root is the reciprocal of the other, we must have \(a = c\).
\(\Rightarrow k - 1 = 3\)
\(\Rightarrow k = 3 + 1\)
\(\Rightarrow k = 4\)
Therefore, Assertion (A) is true.
Correct Option: (a)
💡 Important Points to Remember
✔ Always write the equation in standard form
✔ Carefully identify a, b, and c
✔ Calculate discriminant correctly
✔ Check the sign of D properly
✔ Write conclusions clearly in exams
🎯 Board Exam Importance
📌 Very important for SEBA/HSLC Board Exams
📌 Frequently asked in short and long answer questions
📌 Easy scoring topic if formulas are remembered properly
📚 Why Exercise 4.4 is Important?
This exercise helps students:
✅ Understand the behavior of quadratic equations
✅ Learn how roots change with different discriminant values
✅ Build strong algebraic concepts
✅ Improve problem-solving skills for higher mathematics
🌐 Complete Exercise 4.4 Solutions
👉 Get all question answers with detailed explanations only on:
📣 Stay Connected with Digital Pipal Academy
📺 YouTube: Digital Pipal Academy
📸 Instagram: @pipalacademy
❤️ Final Words
Exercise 4.4 is very easy once you understand the concept of the discriminant. Practice a few problems daily and you will be able to identify the nature of roots within seconds.
📘 Concept + Practice + Revision = Full Marks in Mathematics 💯
Class 10 Maths Chapter 4 – Quadratic Equations (Exercise 4.4) FAQs
Exercise 4.4 teaches how to solve quadratic equations using the Quadratic Formula.
The Quadratic Formula is used to find the roots of a quadratic equation of the form ax² + bx + c = 0.
::contentReference[oaicite:0]{index=0}Exercise 4.4 is very important for HSLC examinations because direct formula-based questions are frequently asked.
The Quadratic Formula is mainly used when factorization or completing the square method becomes difficult.
The discriminant is the value inside the square root part of the quadratic formula.
:contentReference[oaicite:1]{index=1}It helps determine the nature of the roots.
Students often make mistakes while substituting values of a, b, and c or during simplification of square roots.
Students should practice formula-based sums regularly and revise important concepts and examples daily.
Digital Pipal Academy provides easy explanations, step-by-step solutions, and exam-oriented preparation for students.
You can find complete step-by-step solutions for Exercise 4.4 on Digital Pipal Academy.
.png)
