Samacheer Kalvi 10th Maths Exercise 3.10 Solutions

10th Maths Exercise 3.10 solutions (Chapter 3 Algebra) with step-by-step answers for all 6 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance6 answers · full steps below

Q1 (i) $x = 2,\ -\dfrac{1}{4}$
Q1 (ii) $p = \dfrac{9}{2},\ -2$
Q1 (iii) $a = 9,\ -2$
Q1 (iv) $x = \dfrac{-5\sqrt{2}}{2},\ -\sqrt{2}$
Q1 (v) $x = \dfrac{1}{4},\ \dfrac{1}{4}$
Q2 $6$ teams

Question 1

(i)

Solve the following quadratic equations by factorization method: $4x^2 - 7x - 2 = 0$

Show steps (7)

$4x^2-7x-2=0$
Product of coefficient of $x^2$ and constant $=4\times(-2)=-8$ .
Split $-7x$ as $-8x+x$ (since $-8\times 1=-8$ ).
$4x^2-8x+x-2=0$
$4x(x-2)+1(x-2)=0$
$(x-2)(4x+1)=0$
$x-2=0$ or $4x+1=0$

Answer

$x = 2,\ -\dfrac{1}{4}$

(ii)

Solve the following quadratic equations by factorization method: $3(p^2 - 6) = p(p + 5)$

Show steps (7)

$3(p^2-6)=p(p+5)$
$3p^2-18=p^2+5p$
$2p^2-5p-18=0$
Product $=2\times(-18)=-36$ ; split $-5p$ as $-9p+4p$ .
$2p^2-9p+4p-18=0$
$p(2p-9)+2(2p-9)=0$
$(2p-9)(p+2)=0$

Answer

$p = \dfrac{9}{2},\ -2$

(iii)

Solve the following quadratic equations by factorization method: $\sqrt{a(a - 7)} = 3\sqrt{2}$

Show steps (7)

$\sqrt{a(a-7)}=3\sqrt{2}$
Squaring both sides, $a(a-7)=18$
$a^2-7a-18=0$
Split $-7a$ as $-9a+2a$ .
$a^2-9a+2a-18=0$
$a(a-9)+2(a-9)=0$
$(a-9)(a+2)=0$

Answer

$a = 9,\ -2$

(iv)

Solve the following quadratic equations by factorization method: $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$

Show steps (7)

$\sqrt{2}\,x^2+7x+5\sqrt{2}=0$
Product $=\sqrt{2}\times 5\sqrt{2}=10$ ; split $7x$ as $5x+2x$ .
$\sqrt{2}x^2+5x+2x+5\sqrt{2}=0$
$x(\sqrt{2}x+5)+\sqrt{2}(\sqrt{2}x+5)=0$
$(\sqrt{2}x+5)(x+\sqrt{2})=0$
$\sqrt{2}x+5=0\Rightarrow x=\dfrac{-5}{\sqrt{2}}=\dfrac{-5\sqrt{2}}{2}$
$x+\sqrt{2}=0\Rightarrow x=-\sqrt{2}$

Answer

$x = \dfrac{-5\sqrt{2}}{2},\ -\sqrt{2}$

(v)

Solve the following quadratic equations by factorization method: $2x^2 - x + \dfrac{1}{8} = 0$

Show steps (6)

$2x^2-x+\dfrac{1}{8}=0$
Multiplying by $8$ : $16x^2-8x+1=0$
Split $-8x$ as $-4x-4x$ (product $=16\times1=16$ ).
$16x^2-4x-4x+1=0$
$4x(4x-1)-1(4x-1)=0$
$(4x-1)(4x-1)=0$

Answer

$x = \dfrac{1}{4},\ \dfrac{1}{4}$

Q2. The number of volleyball games that must be scheduled in a league …

The number of volleyball games that must be scheduled in a league with $n$ teams is given by $G(n) = \dfrac{n^2 - n}{2}$ where each team plays with every other team exactly once. A league schedules $15$ games. How many teams are in the league?

Show steps (7)

Given $G(n)=\dfrac{n^2-n}{2}=15$
$n^2-n=30$
$n^2-n-30=0$
$n^2-6n+5n-30=0$
$n(n-6)+5(n-6)=0$
$(n-6)(n+5)=0$
$n=6$ or $n=-5$ ; number of teams cannot be negative, so $n=6$ .

Answer

$6$ teams

Quick check

Find the matrix $X$ if $2X + \begin{pmatrix} 1 & 3 \\ 5 & 7 \end{pmatrix} = \begin{pmatrix} 5 & 7 \\ 9 & 5 \end{pmatrix}$
1. (1) $\begin{pmatrix} -2 & -2 \\ 2 & -1 \end{pmatrix}$
2. (2) $\begin{pmatrix} 2 & 2 \\ 2 & -1 \end{pmatrix}$
3. (3) $\begin{pmatrix} 1 & 2 \\ 2 & 2 \end{pmatrix}$
4. (4) $\begin{pmatrix} 2 & 1 \\ 2 & 2 \end{pmatrix}$
Show answer
Answer: (2) $\begin{pmatrix} 2 & 2 \\ 2 & -1 \end{pmatrix}$
1. $2X = \begin{pmatrix} 5 & 7 \\ 9 & 5 \end{pmatrix} - \begin{pmatrix} 1 & 3 \\ 5 & 7 \end{pmatrix} = \begin{pmatrix} 4 & 4 \\ 4 & -2 \end{pmatrix}$
2. $X = \dfrac{1}{2}\begin{pmatrix} 4 & 4 \\ 4 & -2 \end{pmatrix} = \begin{pmatrix} 2 & 2 \\ 2 & -1 \end{pmatrix}$
Question 1 of 5
The values of $a$ and $b$ if $4x^4 - 24x^3 + 76x^2 + ax + b$ is a perfect square are
1. (1) $100,120$
2. (2) $10,12$
3. (3) $-120,100$
4. (4) $12,10$
Show answer
Answer: (3) $-120,100$
1. Let $4x^4 - 24x^3 + 76x^2 + ax + b = (2x^2 - 6x + c)^2$ .
2. Expanding: $4x^4 - 24x^3 + (36 + 4c)x^2 - 12cx + c^2$ .
3. Compare $x^2$ terms: $36 + 4c = 76 \Rightarrow c = 10$ .
4. So $a = -12(10) = -120$ and $b = 10^2 = 100$ .
Question 2 of 5
The solution of the system $x + y - 3z = -6$ , $-7y + 7z = 7$ , $3z = 9$ is
1. (1) $x = 1,\ y = 2,\ z = 3$
2. (2) $x = -1,\ y = 2,\ z = 3$
3. (3) $x = -1,\ y = -2,\ z = 3$
4. (4) $x = 1,\ y = -2,\ z = 3$
Show answer
Answer: (1) $x = 1,\ y = 2,\ z = 3$
1. From $3z = 9$ , $z = 3$ .
2. Put $z=3$ in $-7y + 7z = 7$ : $-7y + 21 = 7 \Rightarrow y = 2$ .
3. Put $y=2,\ z=3$ in $x + y - 3z = -6$ : $x + 2 - 9 = -6 \Rightarrow x = 1$ .
Question 3 of 5
If $A$ is a $2 \times 3$ matrix and $B$ is a $3 \times 4$ matrix, how many columns does $AB$ have
1. (1) $3$
2. (2) $4$
3. (3) $2$
4. (4) $5$
Show answer
Answer: (2) $4$
1. $A_{2\times 3} B_{3\times 4}$ gives order $2 \times 4$ .
2. So $AB$ has $4$ columns.
Question 4 of 5
The square root of $\dfrac{256 x^8 y^4 z^{10}}{25 x^6 y^6 z^6}$ is equal to
1. (1) $\dfrac{16}{5}\left|\dfrac{x^2 z^4}{y^2}\right|$
2. (2) $16\left|\dfrac{y^2}{x^2 z^4}\right|$
3. (3) $\dfrac{16}{5}\left|\dfrac{y}{xz^2}\right|$
4. (4) $\dfrac{16}{5}\left|\dfrac{xz^2}{y}\right|$
Show answer
Answer: (4) $\dfrac{16}{5}\left|\dfrac{xz^2}{y}\right|$
1. $\dfrac{256x^8y^4z^{10}}{25x^6y^6z^6} = \dfrac{256\,x^2 z^4}{25\,y^2}$
2. Taking square root: $\dfrac{16}{5}\left|\dfrac{x z^2}{y}\right|$
Question 5 of 5

Chapter 3 formulas

Roots of a quadratic equation
$ax^2 + bx + c = 0,\; (a \neq 0) \;\Rightarrow\; x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Sum of the roots
$\alpha + \beta = \dfrac{-b}{a} = \dfrac{-\text{Co-efficient of } x}{\text{Co-efficient of } x^2}$
Product of the roots
$\alpha\beta = \dfrac{c}{a} = \dfrac{\text{Constant term}}{\text{Co-efficient of } x^2}$
Quadratic equation formed from its roots
$x^2 - (\alpha + \beta)x + \alpha\beta = 0$
Discriminant
$\Delta = b^2 - 4ac$
Nature of roots: real and unequal
$\Delta > 0 \;\Rightarrow\; \text{the roots are real and unequal}$
Nature of roots: real and equal
$\Delta = 0 \;\Rightarrow\; \text{the roots are real and equal}$
Nature of roots: no real roots
$\Delta < 0 \;\Rightarrow\; \text{there are no real roots}$
Cases for a system of linear equations in three variables
$\text{(i) Unique solution}\quad \text{(ii) Infinitely many solutions}\quad \text{(iii) No solution}$
LCM of algebraic expressions
$\text{LCM} = \text{expression of lowest degree that each given expression exactly divides}$
Quadratic polynomial and its zeroes
$p(x) = ax^2 + bx + c \;(a \neq 0) \;\Rightarrow\; p(x) \text{ has atmost two zeroes, which lie on the } x\text{-axis}$
Order of a matrix
$\text{Order of } A = (\text{Number of rows}) \times (\text{Number of columns}) = m \times n$
Row matrix (row vector)
$A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \end{pmatrix}_{1 \times n}$
Column matrix (column vector)
$A = \begin{pmatrix} a_{11} \\ a_{21} \\ \vdots \\ a_{m1} \end{pmatrix}_{m \times 1}$
Square matrix
$\text{Number of rows} = \text{Number of columns},\; \text{order } n \times n$
Zero matrix (null matrix)
$a_{ij} = 0 \text{ for all } i,\, j$
Transpose of a matrix
$\left(A^{T}\right)_{ij} = a_{ji} \quad (\text{rows and columns of } A \text{ interchanged})$
Diagonal matrix
$a_{ij} = 0 \text{ for } i \neq j \quad (\text{square matrix})$
Scalar matrix
$a_{ij} = \begin{cases} k, & i = j \\ 0, & i \neq j \end{cases}$
Identity matrix (unit matrix)
$a_{ij} = \begin{cases} 1, & i = j \\ 0, & i \neq j \end{cases}$
Lower triangular matrix
$a_{ij} = 0 \text{ for } i < j \quad (\text{all entries above the leading diagonal are zero})$
Upper triangular matrix
$a_{ij} = 0 \text{ for } i > j \quad (\text{all entries below the leading diagonal are zero})$
Equality of matrices
$A = B \iff \text{same order and } a_{ij} = b_{ij} \text{ for all } i,\, j$

Answers checked against the Samacheer Kalvi textbook · Updated 28 September 2026

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