Class 10 Maths Chapter 4 Exercise 4.4 Solutions English Medium | Quadratic Equations SEBA 2026–2027

📘 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 

1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them.

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: \(\sqrt{3}, \frac{\sqrt{3}}{2}\)

 

2. Find the values of \(k\) for each of the following quadratic equations, so that they have two equal real roots.

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

 

New 

💡 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:

www.pipalacademy.com

📣 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.

👉 Students learn to apply the formula directly to find the roots of quadratic equations.

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.

🎯 Correct calculation and proper steps can help students score full marks.

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.

🚀 Pro Tip: Always check the signs of the coefficients carefully before calculation.

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.

⭐ It helps students build confidence and solve Maths problems faster.

You can find complete step-by-step solutions for Exercise 4.4 on Digital Pipal Academy.