Exercise 2.4 Class 10 Solutions Complete Guide
In this article, we explain the complete solutions of Exercise 2.4 Class 10, which deals with symmetric functions of the roots of a quadratic equation. This topic helps you understand how different expressions involving the roots can be simplified using their sum and product. The solutions are written in simple steps and follow the methods taught in the Punjab Textbook Board syllabus, so you can learn the concepts easily and prepare confidently for your exams.
Guide for Solving Question 1.
Before solving this question, remember that we do not need the actual values of α and β. Instead, we only use the formulas for the sum and product of roots:
- Sum of roots: α + β = −b / a
- Product of roots: α × β = c / a
These two values allow us to simplify expressions without finding the roots. We also use the identity: (α + β)² = α² + β² + 2 × α × β
Using these formulas makes it easy to evaluate expressions such as: α² + β², α³β + αβ³, and
α / β + β / α.
Question 1 of Exercise 2.4 Class 10.
If α, β are the roots of the equation x² + px + q = 0, then evaluate:
Solution:
a = 1, b = p, c = q
Sum of Roots, α + β = −p
Product of Roots, αβ = q
(i) α² + β²
(α + β)² = α² + β² + 2αβ
α² + β² = (α + β)² − 2αβ
= (−p)² − 2q
= p² − 2q
(ii) α³β + αβ³
α³β + αβ³ = αβ(α² + β²)
= q[(α + β)² − 2αβ]
= q[(−p)² − 2q]
= q(p² − 2q)
(iii) α/β + β/α
α/β + β/α = (α² + β²) / (αβ)
= (1 / αβ) · (α² + β²)
= (1 / q) · [(α + β)² − 2αβ]
= (1 / q) · [ (−p)² − 2q ]
= (1 / q) · (p² − 2q)
Guide for Solving Question 2.
Before solving this question, note that we now have actual numbers for a, b, and c in the quadratic equation. We still use the sum and product of roots formulas:
- Sum of roots: α + β = −b / a
- Product of roots: α × β = c / a
These values let us calculate expressions like 1/α + 1/β, α²β², 1/(α²β) + 1/(αβ²), and α²/β + β²/α without finding the roots individually.
We also apply standard identities like:
- 1/α + 1/β = (α + β) / (α × β)
- α³ + β³ = (α + β) × ((α + β)² − 3 × α × β)
This approach simplifies calculations and avoids mistakes.
Question 2 of Exercise 2.4 Class 10.
\(\quad \text{If } \alpha, \beta \text{ are the roots of the equation } 4x^2 – 5x + 6 = 0, \text{ then find the values of}\)
\((i) \quad \frac{1}{\alpha} + \frac{1}{\beta} \quad (ii) \quad \alpha^2\beta^2 \quad (iii) \quad \frac{1}{\alpha^2 \beta} + \frac{1}{\alpha \beta^2} \quad (iv) \quad \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha}\)
\begin{align*}
& \textbf{Solution:} \\
& \quad a = 4, \, b = -5, \, c = 6 \\
& \text{Sum of roots,} \\
& \alpha + \beta = \frac{-b}{a} = \frac{-(-5)}{4} = \frac{5}{4} \\
& \text{Product of roots,} \\
& \alpha\beta = \frac{c}{a} = \frac{6}{4} = \frac{3}{2} \\
\end{align*}
\begin{align*}
\text{(i)} \quad \frac{1}{\alpha} + \frac{1}{\beta} \\
&= \frac{\alpha + \beta}{\alpha \beta} \\
&= \frac{\frac{5}{4}}{\frac{3}{2}} \\
&= \frac{5}{4} \times \frac{2}{3} \\
&= \frac{5}{6}
\end{align*}
\begin{align*}
\text{(ii)} \quad \alpha^2\beta^2 \\
&= (\alpha\beta)^2 \\
&= \left( \frac{3}{2} \right)^2 \\
&= \frac{9}{4}
\end{align*}
\(\text{(iii)} \quad \frac{1}{\alpha^2 \beta} + \frac{1}{\alpha \beta^2}\)
\begin{align*}
\\
&= \frac{\alpha + \beta}{(\alpha\beta)^2} \\
&= \frac{\frac{5}{4}}{\left( \frac{3}{2} \right)^2} \\
&= \frac{\frac{5}{4}}{\frac{9}{4}} \\
&= \frac{5}{9}
\end{align*}
\(\text{(iv)} \quad \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha}\)
\begin{align*}
\\
&= \frac{\alpha^3 + \beta^3}{\alpha \beta} \\
\alpha^3 + \beta^3 &= (\alpha + \beta) \left[ (\alpha + \beta)^2 – 3\alpha \beta \right] \\
&= \frac{5}{4} \left[ \left( \frac{5}{4} \right)^2 – 3 \times \frac{3}{2} \right] \\
&= \frac{5}{4} \left[ \frac{25}{16} – \frac{9}{2} \right] \\
&= \frac{5}{4} \left[ \frac{25}{16} – \frac{72}{16} \right] \\
&= \frac{5}{4} \times \left( \frac{-47}{16} \right) \\
&= \frac{5 \times -47}{4 \times 16} \\
&= \frac{-235}{64} \\
\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} &= \frac{-235}{64} \times \frac{2}{3} \\
&= \frac{-470}{192} \\
&= \frac{-235}{96}
\end{align*}
Guide for Solving Question 3.
Before solving this question, observe that the expressions now involve higher powers of α and β. We still rely on the sum and product of roots:
- Sum of roots: α + β = −m / l
- Product of roots: α × β = n / l
Using these two values, we can evaluate expressions like α³β² + α²β³ and 1/α² + 1/β².
We also use the following identities to simplify calculations:
- α³β² + α²β³ = (α × β)² × (α + β)
- 1/α² + 1/β² = (α + β)² − 2 × α × β / (α × β)²
This method allows us to solve these expressions quickly and correctly without computing the roots individually.
Question 3 of Exercise 2.4 Class 10.
\( \text{If } \alpha, \beta \text{ are the roots of the equation } lx^2 + mx + n = 0 \, (l \neq 0) \text{ then find the value of}\)
\((i) \quad \alpha^3\beta^2 + \alpha^2\beta^3 \quad (ii) \quad \frac{1}{\alpha^2} + \frac{1}{\beta^2}\)
\(\textbf{Solution:} \)
\(\quad lx^2 + mx + n = 0 \)
\(\quad ax^2 + bx + c = 0 \)
\(\quad a = l, \, b = m, \, c = n \)
\(\text{If } \alpha, \beta \text{ be the roots of the given equation:} \)
\(\text{Sum of roots:} \)
\(\alpha + \beta = \frac{-b}{a} = \frac{-m}{l} \)
\(\text{Product of roots:} \)
\(\alpha \beta = \frac{c}{a} = \frac{n}{l} \)
\begin{align*}
\text{(i)} \quad \alpha^3 \beta^2 + \alpha^2 \beta^3 \\
&= \alpha^2 \beta^2 (\alpha + \beta) \\
&= (\alpha \beta)^2 (\alpha + \beta) \\
&= \left( \frac{n}{l} \right)^2 \left( \frac{-m}{l} \right) \\
&= \frac{n^2}{l^2} \times \frac{-m}{l} \\
&= \frac{-mn^2}{l^3}
\end{align*}
\begin{align*}
\text{(ii)} \quad \frac{1}{\alpha^2} + \frac{1}{\beta^2} \\
&= \frac{\alpha^2 + \beta^2}{\alpha^2 \beta^2} \\
&= \frac{(\alpha + \beta)^2 – 2\alpha \beta}{(\alpha \beta)^2} \\
&= \frac{\left( \frac{-m}{l} \right)^2 – 2 \times \frac{n}{l}}{\left( \frac{n}{l} \right)^2} \\
&= \frac{\frac{m^2}{l^2} – 2 \times \frac{n}{l}}{\frac{n^2}{l^2}} \\
&= \frac{m^2 – 2nl}{n^2} \\
&= \frac{1}{n^2}(m^2 – 2nl)
\end{align*}
We hope these solutions to Exercise 2.4 class 10 have helped clarify the concept of symmetric functions of the roots of quadratic equations. If you’d like to review earlier topics, feel free to check out the solutions for Exercise 2.3 on the sum and product of roots. You can also move ahead to Exercise 2.5 for more practice with quadratic equations. Keep practicing to strengthen your math skills!
Cuemath has interesting explanation of these concepts.
Video Lecture
If you’re feeling stuck or need further clarity, no problem! Check out this comprehensive video lecture for Exercise 2.4 class 10, which breaks down each solution in an easy-to-follow manner, helping you fully grasp the concepts before tackling the next exercise.
