Exercise 2.4 Class 10 Solutions Complete Guide

In this post, we provide detailed solutions for Exercise 2.4 Class 10 math, which focuses on the symmetric functions of the roots of quadratic equations. This exercise is essential for understanding how different relationships between the roots can be expressed using symmetric functions. Our solutions are designed to help you grasp the concepts easily and improve your problem-solving skills. Let’s dive into solving these questions step by step!

Question 1 of Exercise 2.4 Class 10.

\(\text{If } \alpha, \beta \text{ are the roots of the equation } x^2 + px + q = 0 \text{ then evaluate}\)

\(\textbf{Solution:}\)
\(a = 1, \, b = p, \, c = q \)

\(\text{Sum of roots,} \)
\(\alpha + \beta = \frac{-b}{a} = \frac{-p}{1} = -p \)
\(\text{Product of roots,} \)
\(\alpha\beta = \frac{c}{a} = \frac{q}{1} = q \)

\begin{align*}
\text{(i)} \quad \alpha^2 + \beta^2 \\
(\alpha + \beta)^2 &= \alpha^2 + \beta^2 + 2\alpha\beta \\
\alpha^2 + \beta^2 &= (\alpha + \beta)^2 – 2\alpha\beta \\
&= (-p)^2 – 2q \\
&= p^2 – 2q
\end{align*}

\begin{align*}
\text{(ii)} \quad \alpha^3 \beta + \alpha \beta^3 \\
\alpha^3 \beta + \alpha \beta^3 &= \alpha \beta (\alpha^2 + \beta^2) \\
&= q[(\alpha + \beta)^2 – 2\alpha \beta] \\
&= q[(-p)^2 – 2q] \\
&= q(p^2 – 2q)
\end{align*}

\begin{align*}
\text{(iii)} \quad \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \\
\frac{\alpha}{\beta} + \frac{\beta}{\alpha} &= \frac{\alpha^2 + \beta^2}{\alpha \beta} \\
&= \frac{1}{\alpha \beta} (\alpha^2 + \beta^2) \\
&= \frac{1}{q} [(\alpha + \beta)^2 – 2\alpha \beta] \\
&= \frac{1}{q} [(-p)^2 – 2q] \\
&= \frac{1}{q} (p^2 – 2q)
\end{align*}

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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*}

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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.