Exercise 2.3 Class 10: Achieve Excellence in Roots and Coefficients
Question 1, Question 2, Question 3, Question 4, Question 5, Question 6.
In this post, we will look at Exercise 2.3 class 10, which is about the roots and coefficients of quadratic equations. We will learn how the roots relate to the coefficients and the important formulas for the sum and product of the roots. Understanding these ideas is important for solving quadratic equations. Below, you will find easy-to-follow solutions and explanations to help you understand these concepts better.
Question 1 of Exercise 2.3 Class 10. Without solving, Find the sum and product of the Roots of the following Quadratic Equations.
(i) \(x^2 – 5x + 3 = 0\)
\begin{align*}
& ax^2 + bx + c = 0 \\
& a = 1, \, b = -5, \, c = 3 \\
& \text{Sum of roots} = -\frac{b}{a} = -\left(\frac{-5}{1}\right) = 5 \\
& \text{Product of roots} = \frac{c}{a} = \frac{3}{1} = 3 \\
\end{align*}
\(\textbf{(ii)} \quad 3x^2 + 7x – 11 = 0\)
\begin{align*}
& ax^2 + bx + c = 0 \\
& a = 3, \, b = 7, \, c = -11 \\
& \text{Sum of roots} = -\frac{b}{a} = -\left(\frac{7}{3}\right) \\
& \text{Product of roots} = \frac{c}{a} = \frac{-11}{3} \\
\end{align*}
\(\textbf{(iii)} \quad px^2 – qx + r = 0\)
\begin{align*}
& ax^2 + bx + c = 0 \\
& a = p, \, b = -q, \, c = r \\
& \text{Sum of roots} = -\frac{b}{a} = \frac{q}{p} \\
& \text{Product of roots} = \frac{c}{a} = \frac{r}{p} \\
\end{align*}
\(\textbf{(iv)} \quad (a + b)x^2 – ax + b = 0 \)
\begin{align*}
& \text{Solution:} \quad (a + b)x^2 – ax + b = 0 \\
& Ax^2 + Bx + C = 0 \\
& A = a + b, \, B = -a, \, C = b \\
& \text{Sum of roots} = -\frac{B}{A} = \frac{a}{a + b} \\
& \text{Product of roots} = \frac{C}{A} = \frac{b}{a + b} \\
\end{align*}
\(\text{(v)} \quad (\ell + m)x^2 + (m + n)x + n – \ell = 0 \)
\begin{align*}
& ax^2 + bx + c = 0 \\
& a = \ell + m, \, b = m + n, \, c = n – \ell \\
& \text{Sum of roots} = -\frac{b}{a} = -\frac{m + n}{\ell + m} \\
& \text{Product of roots} = \frac{c}{a} = \frac{n – \ell}{\ell + m} \\
\end{align*}
\(\text{(vi)} \quad 7x^2 – 5mx + 9n = 0 \)
\begin{align*}
& \text{Solution:} \quad 7x^2 – 5mx + 9n = 0 \\
& ax^2 + bx + c = 0 \\
& a = 7, \, b = -5m, \, c = 9n \\
& \text{Sum of roots} = -\frac{b}{a} = \frac{5m}{7} \\
& \text{Product of roots} = \frac{c}{a} = \frac{9n}{7} \\
\end{align*}
Question 2 of Exercise 2.3 Class 10. Find the Value of \(k\) if
(i) \(2kx^2 – 3x + 4k = 0 \)
\begin{align*}
& 2kx^2 – 3x + 4k = 0 \\
& ax^2 + bx + c = 0 \\
& a = 2k, \, b = -3, \, c = 4k \\
& \text{Let } \alpha, \beta \text{ be the Roots of the Equation} \\
& \text{Sum of the roots} = \alpha + \beta = -\frac{b}{a} \\
& S = \alpha + \beta = \frac{-(-3)}{2k} = \frac{3}{2k} \\
& \text{Product of the Roots} = \alpha \beta = \frac{c}{a} \\
& P = \alpha \beta = \frac{4k}{2k} = 2 \\
& \text{Given condition: } S = 2P \\
& \frac{3}{2k} = 2(2) \\
& \frac{3}{2k} = 4 \\
& 3 = 8k \\
& k = \frac{3}{8}
\end{align*}
(ii) \(\text{ Sum of the roots of the equation } x^2 + (3k – 7)x + 5k = 0 \text{ is } \frac{3}{2} \text{ times the product of roots.} \)
\begin{align*}
& x^2 + (3k – 7)x + 5k = 0 \\
& ax^2 + bx + c = 0 \\
& a = 1, \, b = 3k – 7, \, c = 5k \\
& \text{Let } \alpha, \beta \text{ be the roots of the given equation.} \\
& \text{Sum of the roots} \\
& S = \alpha + \beta = -\frac{b}{a} \\
& S = -\frac{(3k – 7)}{1} \\
& S = -3k + 7 \\
& \text{Product of the roots} = P = \alpha \beta = \frac{c}{a} \\
& P = \frac{5k}{1} \\
& P = 5k \\
& \text{Given condition} \\
& S = \frac{3}{2} P \\
& -3k + 7 = \frac{3}{2} (5k) \\
& 2(-3k + 7) = 3(5k) \\
& -6k + 14 = 15k \\
& 14 = 15k + 6k \\
& 14 = 21k \\
& k = \frac{14}{21} = \frac{2}{3}
\end{align*}
Question 3 of Exercise 2.3 Class 10. \(\textbf{Find } k \textbf{ if,} \)
(i) \(\text{ Sum of the squares of the roots of the equation } 4kx^2 + 3kx – 8 = 0 \text{ is } 2 \)
\begin{align*}
& 4kx^2 + 3kx – 8 = 0 \\
& ax^2 + bx + c = 0 \\
& a = 4k, \, b = 3k, \, c = -8 \\
& \text{Sum of roots} \\
& \alpha + \beta = -\frac{b}{a} \\
& \alpha + \beta = -\frac{3k}{4k} = -\frac{3}{4} \\
& \text{Product of roots} = \alpha \beta = \frac{c}{a} \\
& \alpha \beta = \frac{-8}{4k} = \frac{-2}{k} \\
\\
& \text{Given that the sum of the squares of the roots is 2, i.e.} \\
& \alpha^2 + \beta^2 = 2 \\
& (\alpha + \beta)^2 – 2\alpha \beta = 2 \\
& \left( \frac{-3}{4} \right)^2 – 2\left( \frac{-2}{k} \right) = 2 \\
& \frac{9}{16} + \frac{4}{k} = 2 \\
& \frac{9k + 64}{16k} = 2 \\
& 9k + 64 = 32k \\
& 64 = 32k – 9k \\
& 23k = 64 \\
& k = \frac{64}{23}
\end{align*}
(ii) \(\text{ Sum of square of the roots of the equation } x^2 – 2kx + (2k + 1) = 0 \text{ is } 6. \)
\begin{align*}
& x^2 – 2kx + (2k + 1) = 0 \\
& ax^2 + bx + c = 0 \\
& a = 1, \, b = -2k, \, c = 2k + 1 \\
& \alpha, \beta \text{ be the roots of the given equation.} \\
& S = \alpha + \beta = -\frac{b}{a} \\
& \alpha + \beta = -\frac{(-2k)}{1} = 2k \\
& \alpha \beta = \frac{c}{a} = 2k + 1 \\
& P = \alpha \beta = 2k + 1 \\
\\
& \text{Given condition: } \alpha^2 + \beta^2 = 6 \\
& \alpha^2 + \beta^2 + 2\alpha \beta – 2\alpha \beta = 6 \\
& (\alpha + \beta)^2 – 2(\alpha \beta) = 6 \\
& (2k)^2 – 2(2k + 1) = 6 \\
& 4k^2 – 4k – 2 = 6 \\
& 4k^2 – 4k – 2 – 6 = 0 \\
& 4k^2 – 4k – 8 = 0 \\
& 4(k^2 – k – 2) = 0 \\
& k^2 – k – 2 = 0 \quad (\because 4 \neq 0) \\
& k^2 – 2k + k – 2 = 0 \\
& k(k – 2) + 1(k – 2) = 0 \\
& (k – 2)(k + 1) = 0 \\
& k – 2 = 0 \quad \text{or} \quad k + 1 = 0 \\
& k = 2 \quad \text{or} \quad k = -1 \\
& \therefore k = -1, 2
\end{align*}
Question 4 of Exercise 2.3 Class 10. Find \(p\) if
(i)\(\textbf{ The roots of the equation } x^2 – x + p^2 = 0 \textbf{ differ by unity.}\)
\begin{align*}
& \text{Solution:} \\
& x^2 – x + p^2 = 0 \\
& ax^2 + bx + c = 0 \\
& a = 1, \, b = -1, \, c = p^2 \\
& \text{Let the roots be } \alpha \text{ and } \alpha – 1. \\
& \text{Sum of roots } = S = \alpha + (\alpha – 1) = -\frac{-1}{1} = 1 \\
& 2\alpha – 1 = 1 \\
& \Rightarrow 2\alpha = 2 \\
& \Rightarrow \alpha = 1 \\
\\
& \text{Product of roots } = P = \alpha(\alpha – 1) = \frac{p^2}{1} = p^2 \\
& \alpha(\alpha – 1) = p^2 \\
& \text{Putting the value of } \alpha = 1 \\
& 1(1 – 1) = p^2 \\
& \Rightarrow p^2 = 0 \\
& \Rightarrow p = 0 \\
\end{align*}
(ii) \(\textbf{ Find } p \textbf{ if the roots of the equation } x^2 + 3x + p – 2 = 0 \text{ differ by 2.} \)
Solution: \(x^2 + 3x + p – 2 = 0 \)
\begin{align*}
& a = 1, \, b = 3, \, c = p – 2 \\
& \text{Let } \alpha \text{ and } \alpha – 2 \text{ be the roots.} \\
& \text{Sum of roots: } \\
& S = \alpha + (\alpha – 2) = -\frac{b}{a} = -\frac{3}{1} = -3 \\
& 2\alpha – 2 = -3 \\
& 2\alpha = -3 + 2 \\
& 2\alpha = -1 \\
& \Rightarrow \alpha = \frac{-1}{2} \\
\\
& \text{Product of roots: } P = \alpha(\alpha – 2) = \frac{c}{a} = p – 2 \\
& \alpha(\alpha – 2) = p – 2 \\
& \text{Putting the value of } \alpha \text{ from equation (i)} \\
& \frac{-1}{2} \left(\frac{-1}{2} – 2\right) = p – 2 \\
& \frac{-1}{2} \left(\frac{-5}{2}\right) = p – 2 \\
& \frac{5}{4} = p – 2 \\
& \Rightarrow p = \frac{5}{4} + 2 \\
& p = \frac{5 + 8}{4} \\
& p = \frac{13}{4}
\end{align*}
Question 5 of Exercise 2.3 Class 10. find \(m\) if
(i)\(\textbf{ The roots of the equation } x^2 – 7x + 3m – 5 = 0 \textbf{ satisfy the relation } 3\alpha + 2\beta = 4.\)
\begin{align*}
& \text{Solution:} \\
& \text{Let } \alpha, \beta \text{ be the roots of the given equation.} \\
& x^2 – 7x + 3m – 5 = 0 \\
& a = 1, \, b = -7, \, c = 3m – 5 \\
& \text{Sum of roots: } \alpha + \beta = -\frac{b}{a} = \frac{-(-7)}{1} = 7 \quad \text{(i)} \\
& \text{Product of roots: } \alpha \beta = \frac{c}{a} = 3m – 5 \quad \text{(ii)} \\
& \text{Since } 3\alpha + 2\beta = 4 \text{ (given)} \quad \text{(iii)} \\
& \text{From equation (i): } \alpha + \beta = 7 \\
& \beta = 7 – \alpha \\
& \text{Put } \beta = 7 – \alpha \text{ in equation (iii)} \\
& 3\alpha + 2(7 – \alpha) = 4 \\
& 3\alpha + 14 – 2\alpha = 4 \\
& \alpha + 14 = 4 \\
& \alpha = 4 – 14 \\
& \alpha = -10 \\
& \text{Put } \alpha = -10 \text{ in } \beta = 7 – \alpha \\
& \beta = 7 – (-10) \\
& \beta = 7 + 10 \\
& \beta = 17 \\
& \text{Put } \alpha = -10 \text{ and } \beta = 17 \text{ in equation (ii)} \\
& \alpha \beta = 3m – 5 \\
& (-10)(17) = 3m – 5 \\
& -170 = 3m – 5 \\
& -170 + 5 = 3m \\
& -165 = 3m \\
& m = \frac{-165}{3} = -55 \\
\end{align*}
(ii)\(\textbf{Find } m \textbf{ if the roots of the equation } x^2 + 7x + 3m – 5 = 0 \textbf{ satisfy the relation } 3\alpha – 2\beta = 4.\)
\begin{align*}
& \text{Solution:} \\
& \text{Let } \alpha, \beta \text{ be the roots of the equation.} \\
& x^2 + 7x + 3m – 5 = 0 \\
& a = 1, \, b = 7, \, c = 3m – 5 \\
& \text{Sum of roots: } \alpha + \beta = -\frac{b}{a} = -\frac{7}{1} = -7 \quad \text{(i)} \\
& \text{Product of roots: } \alpha \beta = 3m – 5 \quad \text{(ii)} \\
& \text{From equation (i): } \alpha + \beta = -7 \\
& \beta = -7 – \alpha \\
& \text{Put } \beta = -7 – \alpha \text{ in equation (iii)} \\
& 3\alpha – 2\beta = 4 \\
& 3\alpha – 2(-7 – \alpha) = 4 \\
& 3\alpha + 14 + 2\alpha = 4 \\
& 5\alpha = 4 – 14 \\
& 5\alpha = -10 \\
& \alpha = \frac{-10}{5} = -2 \\
& \text{Put } \alpha = -2 \text{ in equation (i)} \\
& \alpha + \beta = -7 \\
& -2 + \beta = -7 \\
& \beta = -7 + 2 \\
& \beta = -5 \\
& \text{Now put } \alpha = -2 \text{ and } \beta = -5 \text{ in equation (ii)} \\
& \alpha \beta = 3m – 5 \\
& (-2)(-5) = 3m – 5 \\
& 10 = 3m – 5 \\
& 10 + 5 = 3m \\
& 15 = 3m \\
& m = \frac{15}{3} = 5
\\
\end{align*}
(iii)\(\textbf{ Find } m \textbf{ if the roots of the equation } 3x^2 – 2x + 7m + 2 = 0 \textbf{ satisfy the relation } 7\alpha – 3\beta = 18.\)
\begin{align*}
& \text{Solution: Let } \alpha, \beta \text{ be the roots of the equation.} \\
& 3x^2 – 2x + 7m + 2 = 0 \\
& ax^2 + bx + c = 0 \\
& a = 3, \, b = -2, \, c = 7m + 2 \\
& \text{Sum of roots: } \alpha + \beta = -\frac{b}{a} = \frac{-(-2)}{3} = \frac{2}{3} \quad \text{(i)} \\
& \text{Product of roots: } \alpha \beta = \frac{c}{a} = \frac{7m + 2}{3} \quad \text{(ii)} \\
& 7\alpha – 3\beta = 18 \quad \text{(iii)} \\
& \text{From equation (i)} \\
& \beta = \frac{2}{3} – \alpha \\
& \text{Substitute this value in equation (iii)} \\
& 7\alpha – 3\left( \frac{2}{3} – \alpha \right) = 18 \\
& 7\alpha – 2 + 3\alpha = 18 \\
& 10\alpha = 20 \\
& \alpha = 2 \\
& \text{Putting } \alpha = 2 \text{ in equation (i)} \\
& \alpha + \beta = \frac{2}{3} \\
& 2 + \beta = \frac{2}{3} \\
& \beta = \frac{2}{3} – 2 = \frac{-4}{3} \\
\\
& \text{Now putting the values of } \alpha \text{ and } \beta \text{ in equation (ii)} \\
& \alpha \beta = \frac{7m + 2}{3} \\
& 2\left( \frac{-4}{3} \right) = \frac{7m + 2}{3} \\
& \frac{-8}{3} = \frac{7m + 2}{3} \\
& -8 = 7m + 2 \\
& 7m = -8 – 2 \\
& m = \frac{-10}{7}
\end{align*}
Question 6 of Exercise 2.3 Class 10.
Find m if sum and product of the roots of the following equations is equal to the given number \(\lambda. \)
(i) \(\boldsymbol{(2m + 3)x^2 + (7m – 5)x + (3m – 10) = 0}\)
\begin{align*}
& \text{Solution: } \\
& (2m + 3)x^2 + (7m – 5)x + (3m – 10) = 0 \\
& ax^2 + bx + c = 0 \\
& a = 2m + 3, \, b = 7m – 5, \, c = 3m – 10 \\
& \text{Let } \alpha, \beta \text{ be the roots of the given equation.} \\
& \text{Sum of Roots: } \alpha + \beta = -\frac{b}{a} = \frac{-(7m – 5)}{2m + 3} = \frac{5 – 7m}{2m + 3} \quad \text{(i)} \\
& \text{Product of Roots: } \alpha \beta = \frac{c}{a} = \frac{3m – 10}{2m + 3} \quad \text{(ii)} \\
& \text{As given, } \alpha + \beta = \lambda \quad \text{(i)} \text{ and } \alpha \beta = \lambda \quad \text{(ii)} \\
& \text{From (i) and (ii):} \\
& \frac{5 – 7m}{2m + 3} = \frac{3m – 10}{2m + 3} \\
& 5 – 7m = 3m – 10 \\
& 5 + 10 = 3m + 7m \\
& 15 = 10m \\
& m = \frac{15}{10} = \frac{3}{2}
\end{align*}
(ii) \(\boldsymbol{4x^2 – (3 + 5m)x – (9m – 17) = 0} \)
\begin{align*}
& \text{Let } \alpha, \beta \text{ be the roots of the equation.} \\
& 4x^2 – (3 + 5m)x – (9m – 17) = 0 \\
& ax^2 + bx + c = 0 \\
& a = 4, \, b = -(3 + 5m), \, c = -(9m – 17) \\
\\
& \text{Sum of roots:} \\
& \alpha + \beta = -\frac{b}{a} = -\left( \frac{-(3 + 5m)}{4} \right) = \frac{3 + 5m}{4} \\
\\
& \text{Product of roots:} \\
& \alpha \beta = \frac{c}{a} = \frac{-(9m – 17)}{4} \\
\\
& \text{Let } \alpha + \beta = \lambda \quad \text{(i)} \text{ and } \alpha \beta = \lambda \quad \text{(ii)} \\
\\
& \text{From (i) and (ii):} \\
& \frac{3 + 5m}{4} = \frac{-(9m – 17)}{4} \\
& 3 + 5m = -9m + 17 \\
& 9m + 5m = 17 – 3 \\
& 14m = 14 \\
& m = \frac{14}{14} = 1
\end{align*}
These solutions are especially written for students to help them understand the concepts of exercise 2.3 class 10 better, and now solutions of exercise 2.4 are waiting for you. Here is an informational article that further explains root and coefficients of quadratic equations. If you still have unclear concepts of exercise 2.3 class 10, take a look at this video lecture. If you want to revisit the previous concepts do take a look at solutions of exercise 2.2.