Prepared by Muhammad Tayyab, Subject Specialist Mathematics, Govt Christian High School Daska
📌 Based on National Curriculum 2023 / PECTAA 2026 Syllabus
📖 What's Inside: This exercise covers forming quadratic equations from given roots using \( x^2 - Sx + P = 0 \), finding equations with transformed roots, and determining conditions for roots. Perfect for Punjab Boards exam preparation.
📚 Related Resources – Unit 2: Quadratic Equations & Inequalities
Class 10 Math Unit 2 Exercise 2.3 – Roots & Coefficients: Complete Guide
Exercise 2.3 of Unit 2 introduces the important relationship between roots and coefficients of quadratic equations. Students learn to form quadratic equations from given roots using the formula \( x^2 - Sx + P = 0 \), find equations with transformed roots, and determine conditions on roots.
What You Will Learn
By working through Exercise 2.3, students master: forming quadratic equations from given roots using sum (\( S = \alpha + \beta \)) and product (\( P = \alpha\beta \)), finding equations with transformed roots (roots increased by a constant, doubled, reciprocals, etc.), and determining conditions for roots to be equal, reciprocal, or satisfy other relationships.
Topics Covered in This Exercise
- Forming Quadratic Equations from Roots – using \( x^2 - Sx + P = 0 \)
- Transformed Roots – roots exceeding by 2, double the roots
- Root Transformations – reciprocals, squares, \( 2\alpha - 1/\beta \)
- Conditions on Roots – reciprocal roots (\( a = c \)), equal roots (\( b^2 = 4ac \))
- Proofs – deriving relationships between coefficients of transformed equations
Why Exercise 2.3 Is Important
Exercise 2.3 is crucial for understanding the relationship between roots and coefficients. Board exam papers frequently feature questions on forming quadratic equations from roots and transforming roots. This exercise builds essential skills for advanced algebra.
Punjab Board Preparation
Students preparing for board exams under any of the 10 BISE Punjab boards should prioritise this exercise. Sum and product of roots, and formation of quadratic equations are frequently tested topics. Practising every part of Q1, Q4, and Q9 by hand is highly recommended.
Exam Tips for Roots & Coefficients
- Remember: \( x^2 - Sx + P = 0 \) where \( S = \alpha + \beta \) and \( P = \alpha\beta \).
- For \( ax^2 + bx + c = 0 \): \( S = -\frac{b}{a} \), \( P = \frac{c}{a} \).
- For transformed roots, carefully calculate the new sum and product.
- Check if your equation has integer coefficients by multiplying by the LCM of denominators.
Common Mistakes Students Make
- Forgetting to use the correct sign for sum of roots \( S \).
- Not multiplying by the LCM to clear fractions.
- Confusing sum and product when forming the equation.
- Making algebraic errors when calculating transformed roots.
📖 Exercise 2.3 – Solved Problems
1 Form a quadratic equation whose roots are given below:
(i) \(-4, 9\)
\[
\begin{aligned}
S &= -4 + 9 = 5,\quad P = (-4)(9) = -36 \\
x^2 - Sx + P &= 0 \implies x^2 - 5x - 36 = 0
\end{aligned}
\]
(ii) \(5, -7\)
\[
\begin{aligned}
S &= 5 + (-7) = -2,\quad P = (5)(-7) = -35 \\
x^2 - (-2)x + (-35) &= 0 \implies x^2 + 2x - 35 = 0
\end{aligned}
\]
(iii) \(\frac{-7}{5}, \frac{-6}{5}\)
\[
\begin{aligned}
S &= \frac{-7}{5} + \frac{-6}{5} = \frac{-13}{5},\quad P = \left(\frac{-7}{5}\right)\left(\frac{-6}{5}\right) = \frac{42}{25} \\
x^2 - \left(\frac{-13}{5}\right)x + \frac{42}{25} &= 0 \implies x^2 + \frac{13}{5}x + \frac{42}{25} = 0 \\
\text{Multiplying by 25: } & 25x^2 + 65x + 42 = 0
\end{aligned}
\]
(iv) \(\frac{-3}{2}, \frac{7}{2}\)
\[
\begin{aligned}
S &= \frac{-3+7}{2} = 2,\quad P = \frac{-21}{4} \\
x^2 - 2x - \frac{21}{4} &= 0 \implies 4x^2 - 8x - 21 = 0
\end{aligned}
\]
(v) \(3 + \sqrt{5}, 3 - \sqrt{5}\)
\[
\begin{aligned}
S &= (3+\sqrt{5}) + (3-\sqrt{5}) = 6 \\
P &= (3+\sqrt{5})(3-\sqrt{5}) = 9 - 5 = 4 \\
x^2 - 6x + 4 &= 0
\end{aligned}
\]
(vi) \(-2 + \sqrt{3}, -2 - \sqrt{3}\)
\[
\begin{aligned}
S &= (-2+\sqrt{3}) + (-2-\sqrt{3}) = -4 \\
P &= (-2+\sqrt{3})(-2-\sqrt{3}) = 4 - 3 = 1 \\
x^2 - (-4)x + 1 &= 0 \implies x^2 + 4x + 1 = 0
\end{aligned}
\]
2 Find the quadratic equation with roots exceeding by 2 than those of roots of \(x^2 + 9x + 20 = 0\)
\[
\begin{aligned}
x^2 + 9x + 20 &= 0 \quad (a=1, b=9, c=20) \\
\alpha + \beta &= -9,\quad \alpha\beta = 20 \\
\text{New roots: } \alpha+2,\; \beta+2 \\
S_{\text{new}} &= (\alpha+2)+(\beta+2) = \alpha+\beta+4 = -9+4 = -5 \\
P_{\text{new}} &= (\alpha+2)(\beta+2) = \alpha\beta + 2(\alpha+\beta) + 4 = 20 + 2(-9) + 4 = 6 \\
x^2 - (-5)x + 6 &= 0 \implies x^2 + 5x + 6 = 0
\end{aligned}
\]
3 Find the equation whose roots are double the roots of \(x^2 - px + q = 0\)
\[
\begin{aligned}
x^2 - px + q &= 0 \quad (a=1, b=-p, c=q) \\
\alpha + \beta &= p,\quad \alpha\beta = q \\
\text{New roots: } 2\alpha,\; 2\beta \\
S_{\text{new}} &= 2\alpha + 2\beta = 2(\alpha+\beta) = 2p \\
P_{\text{new}} &= (2\alpha)(2\beta) = 4\alpha\beta = 4q \\
x^2 - 2px + 4q &= 0
\end{aligned}
\]
4 If \(\alpha,\beta\) are roots of \(x^2 + 2x + 4 = 0\), find the equation whose roots are:
(i) \(\frac{1}{\alpha}, \frac{1}{\beta}\)
\[
\begin{aligned}
\alpha+\beta &= -2,\quad \alpha\beta = 4 \\
S &= \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{-2}{4} = -\frac{1}{2} \\
P &= \frac{1}{\alpha\beta} = \frac{1}{4} \\
x^2 - \left(-\frac{1}{2}\right)x + \frac{1}{4} &= 0 \implies 4x^2 + 2x + 1 = 0
\end{aligned}
\]
(ii) \(\frac{\alpha}{\beta}, \frac{\beta}{\alpha}\)
\[
\begin{aligned}
S &= \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2+\beta^2}{\alpha\beta} = \frac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta} = \frac{4 - 8}{4} = -1 \\
P &= 1 \\
x^2 - (-1)x + 1 &= 0 \implies x^2 + x + 1 = 0
\end{aligned}
\]
(iii) \(2\alpha - \frac{1}{\beta}, 2\beta - \frac{1}{\alpha}\)
\[
\begin{aligned}
S &= 2(\alpha+\beta) - \left(\frac{1}{\alpha}+\frac{1}{\beta}\right) = 2(-2) - \left(\frac{-2}{4}\right) = -4 + \frac{1}{2} = -\frac{7}{2} \\
P &= 4\alpha\beta - 2 - 2 + \frac{1}{\alpha\beta} = 4(4) - 4 + \frac{1}{4} = \frac{49}{4} \\
x^2 - \left(-\frac{7}{2}\right)x + \frac{49}{4} &= 0 \implies 4x^2 + 14x + 49 = 0
\end{aligned}
\]
(iv) \(\alpha^2, \beta^2\)
\[
\begin{aligned}
S &= \alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = 4 - 8 = -4 \\
P &= \alpha^2\beta^2 = (\alpha\beta)^2 = 16 \\
x^2 - (-4)x + 16 &= 0 \implies x^2 + 4x + 16 = 0
\end{aligned}
\]
(v) \(2\alpha - 1, 2\beta - 1\)
\[
\begin{aligned}
S &= (2\alpha-1)+(2\beta-1) = 2(\alpha+\beta) - 2 = -4 - 2 = -6 \\
P &= (2\alpha-1)(2\beta-1) = 4\alpha\beta - 2(\alpha+\beta) + 1 = 16 + 4 + 1 = 21 \\
x^2 - (-6)x + 21 &= 0 \implies x^2 + 6x + 21 = 0
\end{aligned}
\]
5 Find the condition that roots of \(ax^2 + bx + c = 0\) should be reciprocals of each other.
\[
\begin{aligned}
ax^2 + bx + c &= 0 \\
\text{Let one root be } \alpha,\text{ then other is } \frac{1}{\alpha} \\
\alpha \cdot \frac{1}{\alpha} &= \frac{c}{a} \implies 1 = \frac{c}{a} \implies a = c
\end{aligned}
\]
✅ Condition: \(a = c\)
6 Find the value of \(k\), given that one root of \(x^2 - (2k+4)x + (7k+1) = 0\) is \(3\).
\[
\begin{aligned}
x^2 - (2k+4)x + (7k+1) &= 0 \\
\text{Since } 3 \text{ is a root, put } x &= 3: \\
9 - (2k+4)(3) + (7k+1) &= 0 \\
9 - 6k - 12 + 7k + 1 &= 0 \\
k - 2 &= 0 \implies k = 2
\end{aligned}
\]
✅ \(k = 2\)
7 Find the value of \(m\) in \(2x^2 + 3x + m = 0\) when sum of its roots is equal to double the product of its roots.
\[
\begin{aligned}
2x^2 + 3x + m &= 0 \quad (a=2, b=3, c=m) \\
\alpha + \beta &= -\frac{3}{2},\quad \alpha\beta = \frac{m}{2} \\
\text{Given: } \alpha + \beta &= 2(\alpha\beta) \\
-\frac{3}{2} &= 2\left(\frac{m}{2}\right) \implies -\frac{3}{2} = m
\end{aligned}
\]
✅ \(m = -\frac{3}{2}\)
8 If \(\alpha,\beta\) are roots of \(x^2 + ax + b = 0\) and \(\alpha^2,\beta^2\) are roots of \(x^2 + Ax + B = 0\), prove that \(A = 2b - a^2,\; B = b^2\).
\[
\begin{aligned}
x^2 + ax + b &= 0 \implies \alpha+\beta = -a,\quad \alpha\beta = b \\
\text{For } x^2 + Ax + B &= 0,\text{ roots are } \alpha^2,\beta^2: \\
\alpha^2 + \beta^2 &= (\alpha+\beta)^2 - 2\alpha\beta = (-a)^2 - 2b = a^2 - 2b \\
\text{But } \alpha^2 + \beta^2 &= -A \implies -A = a^2 - 2b \implies A = 2b - a^2 \\
\alpha^2\beta^2 &= (\alpha\beta)^2 = b^2 \implies B = b^2
\end{aligned}
\]
✅ Proved: \(A = 2b - a^2,\; B = b^2\)
9 If \(\alpha,\beta\) are roots of \(x^2 + px + q = 0\), find the condition that:
(i) \(\alpha = \beta\)
\[
\begin{aligned}
\alpha - \beta &= 0 \implies (\alpha-\beta)^2 = 0 \\
(\alpha+\beta)^2 - 4\alpha\beta &= 0 \implies (-p)^2 - 4q = 0 \\
p^2 &= 4q
\end{aligned}
\]
✅ Condition: \(p^2 = 4q\)
(ii) \(\alpha = \frac{1}{\beta}\)
\[
\alpha = \frac{1}{\beta} \implies \alpha\beta = 1 \implies q = 1
\]
✅ Condition: \(q = 1\)
📝 Multiple Choice Questions (Unit 2 Review)
1. The sum of roots of \(x^2 - 5x + 6 = 0\) is:
✅ Correct Answer: (A) 5
For \(x^2 - 5x + 6 = 0\), \(a=1, b=-5\), so \(S = -\frac{b}{a} = 5\).
For \(x^2 - 5x + 6 = 0\), \(a=1, b=-5\), so \(S = -\frac{b}{a} = 5\).
2. The product of roots of \(2x^2 + 3x - 5 = 0\) is:
✅ Correct Answer: (B) \(-\frac{5}{2}\)
For \(2x^2 + 3x - 5 = 0\), \(P = \frac{c}{a} = \frac{-5}{2}\).
For \(2x^2 + 3x - 5 = 0\), \(P = \frac{c}{a} = \frac{-5}{2}\).
3. The quadratic equation with roots \(2\) and \(3\) is:
✅ Correct Answer: (A) \(x^2 - 5x + 6 = 0\)
\(S = 2+3 = 5\), \(P = 6\), so \(x^2 - 5x + 6 = 0\).
\(S = 2+3 = 5\), \(P = 6\), so \(x^2 - 5x + 6 = 0\).
4. If roots are reciprocal of each other, then:
✅ Correct Answer: (B) \(a = c\)
For reciprocal roots, \( \alpha\beta = 1 \implies \frac{c}{a} = 1 \implies a = c\).
For reciprocal roots, \( \alpha\beta = 1 \implies \frac{c}{a} = 1 \implies a = c\).
5. If roots are equal, then:
✅ Correct Answer: (A) \(b^2 = 4ac\)
Equal roots occur when discriminant \(b^2 - 4ac = 0 \implies b^2 = 4ac\).
Equal roots occur when discriminant \(b^2 - 4ac = 0 \implies b^2 = 4ac\).
📈 Key Concepts – Roots & Coefficients
- Quadratic from roots: \(x^2 - Sx + P = 0\), where \(S = \alpha+\beta\), \(P = \alpha\beta\)
- Sum of roots: \(S = -\frac{b}{a}\) for \(ax^2 + bx + c = 0\)
- Product of roots: \(P = \frac{c}{a}\) for \(ax^2 + bx + c = 0\)
- Transformed roots: Calculate new sum and product, then form equation
- Reciprocal roots: \(a = c\)
- Equal roots: \(b^2 = 4ac\)
❓ Frequently Asked Questions
What is taught in Exercise 2.3 of Class 10 Math Unit 2?
Exercise 2.3 covers forming quadratic equations from given roots using the formula \( x^2 - Sx + P = 0 \), where \( S = \alpha + \beta \) (sum of roots) and \( P = \alpha\beta \) (product of roots). It also includes finding equations with transformed roots and conditions on roots.
How many questions are there in Unit 2 Exercise 2.3?
Exercise 2.3 has 9 questions covering formation of quadratic equations from roots (Q1), transformed roots (Q2, Q3, Q4), conditions on roots (Q5, Q6, Q7), and proofs (Q8, Q9).
What is the formula for forming a quadratic equation from roots?
The formula is \( x^2 - Sx + P = 0 \), where \( S \) is the sum of roots (\( \alpha + \beta \)) and \( P \) is the product of roots (\( \alpha\beta \)).
What are the sum and product of roots of \( ax^2+bx+c=0 \)?
For \( ax^2+bx+c=0 \), the sum of roots is \( -\frac{b}{a} \) and the product of roots is \( \frac{c}{a} \).
How do you find an equation with transformed roots?
To find an equation with transformed roots, first find the sum and product of the original roots using \( \alpha+\beta = -\frac{b}{a} \) and \( \alpha\beta = \frac{c}{a} \). Then calculate the sum and product of the transformed roots and use \( x^2 - Sx + P = 0 \).
Is this solution according to the PECTAA 2026 syllabus?
Yes, these solutions are prepared according to the PECTAA 2026 / National Curriculum 2023 syllabus for Class 10 Mathematics.
Is this Exercise 2.3 solution valid for all Punjab Boards?
Yes, the content follows the unified Punjab textbook and is applicable to students of all 10 BISE Punjab boards.
Are solved PDF notes available for Exercise 2.3?
Yes, a complete solved PDF for Exercise 2.3 is embedded on this page and available to download for free.
Can I download the Unit 2 Exercise 2.3 solution as a PDF?
Yes, use the Download PDF button on this page to save the complete solved Exercise 2.3 notes to your device.
Is Exercise 2.3 important for Class 10 board exams?
Yes, sum and product of roots, and formation of quadratic equations are frequently tested in board exams. Exercise 2.3 builds essential skills in understanding the relationship between roots and coefficients.
Who prepared these Class 10 Math Unit 2 notes?
These notes were prepared by Muhammad Tayyab, Subject Specialist Mathematics at Govt Christian High School Daska, for Hira Science Academy.