Welcome to the complete solutions of Exercise 4.3 – Quadratic Equations for Class 10 Mathematics (English Medium).
In this exercise, students learn how to solve quadratic equations using the relationship between the roots and coefficients of a quadratic equation. This is an important concept for board examinations and also helps students build a strong mathematical foundation for higher classes.
At Digital Pipal Academy, our experienced teachers have prepared simple, clear, and exam-oriented solutions so every student can understand the chapter easily.
🧮 Step-by-Step Solutions For Exercise 4.3
General Formula: For an equation \(ax^2 + bx + c = 0\), the roots are given by:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
(i) \(2x^2 - 7x + 3 = 0\)
Here, \(a = 2, b = -7, c = 3\).
Discriminant \(D = b^2 - 4ac = (-7)^2 - 4(2)(3) = 49 - 24 = 25\).
\(\therefore x = \frac{-(-7) \pm \sqrt{25}}{2(2)} = \frac{7 \pm 5}{4}\).
Either \(x = \frac{12}{4} = 3\) or \(x = \frac{2}{4} = \frac{1}{2}\).
Roots: \(3, \frac{1}{2}\)
(ii) \(2x^2 + x - 4 = 0\)
Here, \(a = 2, b = 1, c = -4\).
Discriminant \(D = 1^2 - 4(2)(-4) = 1 + 32 = 33\).
\(\therefore x = \frac{-1 \pm \sqrt{33}}{4}\).
Roots: \(\frac{-1 + \sqrt{33}}{4}, \frac{-1 - \sqrt{33}}{4}\)
(iii) \(4x^2 + 4\sqrt{3}x + 3 = 0\)
Here, \(a = 4, b = 4\sqrt{3}, c = 3\).
Discriminant \(D = (4\sqrt{3})^2 - 4(4)(3) = 48 - 48 = 0\).
\(\therefore x = \frac{-4\sqrt{3} \pm 0}{8} = -\frac{\sqrt{3}}{2}\).
Roots: \(-\frac{\sqrt{3}}{2}, -\frac{\sqrt{3}}{2}\)
(iv) \(2x^2 + 5\sqrt{3}x + 6 = 0\)
Here, \(a = 2, b = 5\sqrt{3}, c = 6\).
Discriminant \(D = (5\sqrt{3})^2 - 4(2)(6) = 75 - 48 = 27\).
\(\therefore x = \frac{-5\sqrt{3} \pm \sqrt{27}}{4} = \frac{-5\sqrt{3} \pm 3\sqrt{3}}{4}\).
Roots: \(-\frac{\sqrt{3}}{2}, -2\sqrt{3}\).
(v) \(x^2 + 4x + 1 = 0\)
Here, \(a = 1, b = 4, c = 1\).
Discriminant \(D = 4^2 - 4(1)(1) = 12\).
\(\therefore x = \frac{-4 \pm 2\sqrt{3}}{2} = -2 \pm \sqrt{3}\).
Roots: \(-2 + \sqrt{3}, -2 - \sqrt{3}\)
(vi) \(4x^2 + x - 3 = 0\)
Here, \(a = 4, b = 1, c = -3\).
Discriminant \(D = 1^2 - 4(4)(-3) = 49\).
\(\therefore x = \frac{-1 \pm 7}{8}\).
Roots: \(\frac{3}{4}, -1\).
(vii) \(abx^2 + (b^2 - ac)x - bc = 0\)
Here, \(a = ab, b = (b^2 - ac), c = -bc\).
Discriminant \(D = (b^2 - ac)^2 - 4(ab)(-bc) = (b^2 + ac)^2\).
\(\therefore x = \frac{-(b^2 - ac) \pm (b^2 + ac)}{2ab}\).
Roots: \(\frac{c}{b}, -\frac{b}{a}\).
(viii) \(x^2 - 2ax + (a^2 - b^2) = 0\)
Here, \(a = 1, b = -2a, c = (a^2 - b^2)\).
Discriminant \(D = (-2a)^2 - 4(a^2 - b^2) = 4b^2\).
\(\therefore x = \frac{2a \pm 2b}{2} = a \pm b\).
Roots: \(a + b, a - b\)
(ix) \(a(x^2 + 1) = x(a^2 + 1)\)
Rearranged: \(ax^2 - (a^2 + 1)x + a = 0\).
Discriminant \(D = (a^2 + 1)^2 - 4(a)(a) = (a^2 - 1)^2\).
\(\therefore x = \frac{(a^2 + 1) \pm (a^2 - 1)}{2a}\).
Roots: \(a, \frac{1}{a}\).
(x) \(8x(2x - 1) + 1 = 0\)
Rearranged: \(16x^2 - 8x + 1 = 0\).
Discriminant \(D = (-8)^2 - 4(16)(1) = 0\).
\(\therefore x = \frac{8 \pm 0}{32} = \frac{1}{4}\).
Roots: \(\frac{1}{4}, \frac{1}{4}\)
Solution:
Given the equation:
\(\dfrac{x-2}{x-3} + \dfrac{x-4}{x-5} = \dfrac{10}{3}\)
Taking the L.C.M. on the left hand side:
\(\dfrac{(x-2)(x-5) + (x-4)(x-3)}{(x-3)(x-5)} = \dfrac{10}{3}\)
\(\Rightarrow \dfrac{(x^2 - 5x - 2x + 10) + (x^2 - 3x - 4x + 12)}{x^2 - 5x - 3x + 15} = \dfrac{10}{3}\)
\(\Rightarrow \dfrac{(x^2 - 7x + 10) + (x^2 - 7x + 12)}{x^2 - 8x + 15} = \dfrac{10}{3}\)
\(\Rightarrow \dfrac{2x^2 - 14x + 22}{x^2 - 8x + 15} = \dfrac{10}{3}\)
By cross-multiplication:
\(3(2x^2 - 14x + 22) = 10(x^2 - 8x + 15)\)
\(\Rightarrow 6x^2 - 42x + 66 = 10x^2 - 80x + 150\)
\(\Rightarrow 10x^2 - 6x^2 - 80x + 42x + 150 - 66 = 0\)
\(\Rightarrow 4x^2 - 38x + 84 = 0\)
Dividing the equation by 2 (Standard Form):
\(2x^2 - 19x + 42 = 0\)
Now, using the factorization method (splitting the middle term):
Here, \(2 \times 42 = 84\). We need two numbers whose product is 84 and sum is 19 (the numbers are 12 and 7).
\(\Rightarrow 2x^2 - 12x - 7x + 42 = 0\)
\(\Rightarrow 2x(x - 6) - 7(x - 6) = 0\)
\(\Rightarrow (x - 6)(2x - 7) = 0\)
Therefore,
Either \(x - 6 = 0 \Rightarrow x = 6\)
Or \(2x - 7 = 0 \Rightarrow x = \dfrac{7}{2}\)
The required roots are: \(6\) and \(\dfrac{7}{2}\).
Solution:
(i) \(x - \frac{1}{x} = 3,\; x \ne 0\)
\(\Rightarrow \frac{x^2 - 1}{x} = 3\)
\(\Rightarrow x^2 - 1 = 3x\)
\(\Rightarrow x^2 - 3x - 1 = 0\)
Here, \(a = 1, b = -3, c = -1\).
Using Quadratic Formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
\(\Rightarrow x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-1)}}{2(1)}\)
\(\Rightarrow x = \frac{3 \pm \sqrt{9 + 4}}{2} = \frac{3 \pm \sqrt{13}}{2}\)
Roots: \(\frac{3 + \sqrt{13}}{2}, \frac{3 - \sqrt{13}}{2}\)
(ii) \(\frac{1}{x+4} - \frac{1}{x-7} = \frac{11}{30},\; x \ne -4, 7\)
\(\Rightarrow \frac{(x-7) - (x+4)}{(x+4)(x-7)} = \frac{11}{30}\)
\(\Rightarrow \frac{x - 7 - x - 4}{x^2 - 7x + 4x - 28} = \frac{11}{30}\)
\(\Rightarrow \frac{-11}{x^2 - 3x - 28} = \frac{11}{30}\)
\(\Rightarrow -1 = \frac{x^2 - 3x - 28}{30}\)
\(\Rightarrow x^2 - 3x - 28 = -30\)
\(\Rightarrow x^2 - 3x + 2 = 0\)
\(\Rightarrow x^2 - 2x - x + 2 = 0\)
\(\Rightarrow x(x-2) - 1(x-2) = 0 \Rightarrow (x-2)(x-1) = 0\)
Roots: \(1, 2\)
(iii) \(\frac{2}{3}x^2 - \frac{1}{3}x - 1 = 0\)
Multiplying the entire equation by 3:
\(\Rightarrow 2x^2 - x - 3 = 0\)
\(\Rightarrow 2x^2 - 3x + 2x - 3 = 0\)
\(\Rightarrow x(2x - 3) + 1(2x - 3) = 0\)
\(\Rightarrow (2x - 3)(x + 1) = 0\)
Roots: \(\frac{3}{2}, -1\)
(iv) \(2x^2 + \frac{1}{2} = 2x\)
Multiplying the entire equation by 2:
\(\Rightarrow 4x^2 + 1 = 4x\)
\(\Rightarrow 4x^2 - 4x + 1 = 0\)
\(\Rightarrow (2x)^2 - 2(2x)(1) + (1)^2 = 0\)
\(\Rightarrow (2x - 1)^2 = 0\)
Roots: \(\frac{1}{2}, \frac{1}{2}\)
(v) \(x + \frac{1}{x} = 2\)
\(\Rightarrow \frac{x^2 + 1}{x} = 2\)
\(\Rightarrow x^2 + 1 = 2x\)
\(\Rightarrow x^2 - 2x + 1 = 0\)
\(\Rightarrow (x - 1)^2 = 0\)
Roots: \(1, 1\)
(vi) \(\frac{5x-6}{4x-1} = \frac{2x+3}{3x+2}\)
By cross-multiplication:
\(\Rightarrow (5x-6)(3x+2) = (2x+3)(4x-1)\)
\(\Rightarrow 15x^2 + 10x - 18x - 12 = 8x^2 - 2x + 12x - 3\)
\(\Rightarrow 15x^2 - 8x - 12 = 8x^2 + 10x - 3\)
\(\Rightarrow 7x^2 - 18x - 9 = 0\)
\(\Rightarrow 7x^2 - 21x + 3x - 9 = 0\)
\(\Rightarrow 7x(x-3) + 3(x-3) = 0 \Rightarrow (x-3)(7x+3) = 0\)
Roots: \(3, -\frac{3}{7}\)
Q4. The sum of the reciprocals of Rehman's ages, 3 years ago and 5 years from now is \(\frac{1}{3}\). Find his present age.
Solution:
Let Rehman's present age be \(x\) years.His age 3 years ago = \((x - 3)\) years.
His age 5 years from now = \((x + 5)\) years.
According to the question,
\(\frac{1}{x-3} + \frac{1}{x+5} = \frac{1}{3}\)
\(\Rightarrow \frac{(x+5) + (x-3)}{(x-3)(x+5)} = \frac{1}{3}\)
\(\Rightarrow \frac{2x + 2}{x^2 + 2x - 15} = \frac{1}{3}\)
\(\Rightarrow 3(2x + 2) = x^2 + 2x - 15\)
\(\Rightarrow 6x + 6 = x^2 + 2x - 15\)
\(\Rightarrow x^2 - 4x - 21 = 0\)
\(\Rightarrow (x - 7)(x + 3) = 0\)
Since age cannot be negative, \(x = 7\).
Answer: Rehman's present age is 7 years.
Q5. In a class test, the sum of Shewali's marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.
Solution:
Let the marks in Mathematics be \(x\).Then, marks in English = \((30 - x)\).
According to the question,
\((x + 2) \times (30 - x - 3) = 210\)
\(\Rightarrow (x + 2)(27 - x) = 210\)
\(\Rightarrow 27x - x^2 + 54 - 2x = 210\)
\(\Rightarrow -x^2 + 25x + 54 - 210 = 0\)
\(\Rightarrow x^2 - 25x + 156 = 0\)
\(\Rightarrow (x - 12)(x - 13) = 0\)
So, \(x = 12\) or \(x = 13\).
Answer: If marks in Math = 12, then English = 18. If marks in Math = 13, then English = 17.
Q6. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Solution:
Let the shorter side be \(x\) metres.Longer side = \((x + 30)\) metres and Diagonal = \((x + 60)\) metres.
By Pythagoras Theorem,
\((\text{Shorter Side})^2 + (\text{Longer Side})^2 = (\text{Diagonal})^2\)
\(\Rightarrow x^2 + (x + 30)^2 = (x + 60)^2\)
\(\Rightarrow x^2 + x^2 + 60x + 900 = x^2 + 120x + 3600\)
\(\Rightarrow x^2 - 60x - 2700 = 0\)
\(\Rightarrow (x - 90)(x + 30) = 0\)
Since side length cannot be negative, \(x = 90\).
Answer: Shorter side is 90 m and Longer side is 120 m.
Q7. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Solution:
Let the larger number be \(x\).The square of the smaller number = \(8x\).
According to the question,
\(x^2 - 8x = 180\)
\(\Rightarrow x^2 - 8x - 180 = 0\)
\(\Rightarrow (x - 18)(x + 10) = 0\)
We take \(x = 18\) (as \(x = -10\) would make the square of the smaller number negative).
Now, Square of smaller number = \(8 \times 18 = 144\).
\(\therefore\) Smaller number = \(\sqrt{144} = \pm 12\).
Answer: The numbers are (18, 12) or (18, -12).
Q8. If \(\alpha\) and \(\beta\) are the roots of the equation \(3x^2 + 8x + 2 = 0\), then the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\) will be—
Solution:
Given equation: \(3x^2 + 8x + 2 = 0\)Here, \(a = 3, b = 8, c = 2\).
We know that:
Sum of roots \((\alpha + \beta) = -\frac{b}{a} = -\frac{8}{3}\)
Product of roots \((\alpha\beta) = \frac{c}{a} = \frac{2}{3}\)
Now, \(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}\)
\(\Rightarrow \frac{-8/3}{2/3} = -4\)
Correct Option: (b) -4
💡 Important Tips for Students
✔ Learn the root formulas properly
✔ Always identify a, b, and c carefully
✔ Don’t forget negative signs
✔ Practice equation formation regularly
✔ Write every step clearly in exams
🎯 Board Exam Importance
📌 Exercise 4.3 is highly important for SEBA/HSLC Board Exams
📌 Questions are often asked directly from root relationships
📌 Formula-based questions are easy scoring if practiced well
🌐 Complete Exercise 4.3 Solutions
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❤️ Final Words
Exercise 4.3 helps students understand the deeper relationship between roots and quadratic equations. Once you understand the formulas and practice regularly, these problems become very easy to solve.
📘 Learn the concept, practice daily, and score full marks in Mathematics.
Class 10 Maths Chapter 4 – Quadratic Equations (Exercise 4.3) FAQs
Exercise 4.3 teaches how to solve quadratic equations using the Completing the Square Method.
It is an algebraic method used to transform a quadratic equation into a perfect square form for easier solving.
Yes, Exercise 4.3 is very important for the HSLC examination and questions of 2–4 marks are often asked from this topic.
This method is used when factorization is difficult or when the equation cannot be solved easily by factoring.
The questions mainly involve converting quadratic equations into square form and finding their roots.
Students often make sign mistakes or errors while adding and subtracting square terms.
Regular practice, learning formulas properly, and solving previous year questions are very important.
Digital Pipal Academy provides easy step-by-step solutions, exam-focused preparation, and updated syllabus-based learning.
You can get the complete step-by-step solution for Exercise 4.3 on Digital Pipal Academy.
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