Samacheer Kalvi 10th Maths Exercise 3.16 Solutions

10th Maths Exercise 3.16 solutions (Chapter 3 Algebra) with step-by-step answers for all 13 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance13 answers · full steps below

Q1 (i)The parabola cuts the $X$ axis at two distinct points; roots are real and unequal, solution set $\{4,5\}$
Q1 (ii)The parabola touches the $X$ axis at one point; roots are real and equal, $x=2,2$
Q1 (iii)The parabola does not meet the $X$ axis; the equation has no real roots
Q1 (iv)The parabola cuts the $X$ axis at two points; roots are real and unequal, solution set $\{-3,3\}$
Q1 (v)The parabola touches the $X$ axis at one point; roots are real and equal, $x=3,3$
Q1 (vi)The parabola cuts the $X$ axis at two points; roots are real and unequal, solution set $\{-2,\ 1.5\}$
Q2Solution set $\{-3,\ 4\}$
Q3The line does not meet the parabola, so $x^2+1=0$ has no real roots
Q4 $x=-1$ (real and equal roots)
Q5Solution set $\{-4,\ 1\}$
Q6Solution set $\{-2,\ 7\}$
Q7Solution set $\{-1,\ 3\}$
Q8Solution set $\{-2,\ 3\}$

Q1. Graph the following quadratic equations and state their nature of solutions.

(i)

$x^2 - 9x + 20 = 0$

Show steps (7)

Let $y=x^2-9x+20$ .
Table of values:
$x$ : $0,\ 1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7$
$y$ : $20,\ 12,\ 6,\ 2,\ 0,\ 0,\ 2,\ 6$
Plot these points and join them by a smooth curve (parabola).
The curve cuts the $X$ axis at two points $(4,0)$ and $(5,0)$ .
So the $x$ coordinates $4$ and $5$ are the solutions.

Answer

The parabola cuts the $X$ axis at two distinct points; roots are real and unequal, solution set $\{4,5\}$

(ii)

$x^2 - 4x + 4 = 0$

Show steps (6)

Let $y=x^2-4x+4$ .
Table of values:
$x$ : $-1,\ 0,\ 1,\ 2,\ 3,\ 4,\ 5$
$y$ : $9,\ 4,\ 1,\ 0,\ 1,\ 4,\ 9$
Plot the points and join them by a smooth curve.
The curve just touches the $X$ axis at one point $(2,0)$ .

Answer

The parabola touches the $X$ axis at one point; roots are real and equal, $x=2,2$

(iii)

$x^2 + x + 7 = 0$

Show steps (6)

Let $y=x^2+x+7$ .
Table of values:
$x$ : $-3,\ -2,\ -1,\ 0,\ 1,\ 2$
$y$ : $13,\ 9,\ 7,\ 7,\ 9,\ 13$
Plot the points and join them by a smooth curve.
The whole curve lies above the $X$ axis, so it neither cuts nor touches the $X$ axis.

Answer

The parabola does not meet the $X$ axis; the equation has no real roots

(iv)

$x^2 - 9 = 0$

Show steps (6)

Let $y=x^2-9$ .
Table of values:
$x$ : $-4,\ -3,\ -2,\ -1,\ 0,\ 1,\ 2,\ 3,\ 4$
$y$ : $7,\ 0,\ -5,\ -8,\ -9,\ -8,\ -5,\ 0,\ 7$
Plot the points and join them by a smooth curve.
The curve cuts the $X$ axis at $(-3,0)$ and $(3,0)$ .

Answer

The parabola cuts the $X$ axis at two points; roots are real and unequal, solution set $\{-3,3\}$

(v)

$x^2 - 6x + 9 = 0$

Show steps (6)

Let $y=x^2-6x+9$ .
Table of values:
$x$ : $0,\ 1,\ 2,\ 3,\ 4,\ 5,\ 6$
$y$ : $9,\ 4,\ 1,\ 0,\ 1,\ 4,\ 9$
Plot the points and join them by a smooth curve.
The curve touches the $X$ axis at only one point $(3,0)$ .

Answer

The parabola touches the $X$ axis at one point; roots are real and equal, $x=3,3$

(vi)

$(2x - 3)(x + 2) = 0$

Show steps (7)

$(2x-3)(x+2)=0$ gives $2x^2+x-6=0$ .
Let $y=2x^2+x-6$ .
Table of values:
$x$ : $-3,\ -2,\ -1,\ 0,\ 1,\ 2,\ 3$
$y$ : $9,\ 0,\ -5,\ -6,\ -3,\ 4,\ 15$
Plot the points and join them by a smooth curve.
The curve cuts the $X$ axis at $(-2,0)$ and $(1.5,0)$ .

Answer

The parabola cuts the $X$ axis at two points; roots are real and unequal, solution set $\{-2,\ 1.5\}$

Q2. Draw the graph of $y = x^2 - 4$ and hence solve $x^2 - x - 12 = 0$

Show steps (12)

Draw $y=x^2-4$ using the table:
$x$ : $-4,\ -3,\ -2,\ -1,\ 0,\ 1,\ 2,\ 3,\ 4$
$y$ : $12,\ 5,\ 0,\ -3,\ -4,\ -3,\ 0,\ 5,\ 12$
To solve $x^2-x-12=0$ , subtract it from $y=x^2-4$ :
$y=x^2-4$
$0=x^2-x-12\quad(-)$
$y=x+8$
This is a straight line. Table for $y=x+8$ :
$x$ : $-4,\ -3,\ 0,\ 2,\ 4$
$y$ : $4,\ 5,\ 8,\ 10,\ 12$
The line cuts the parabola at $(-3,5)$ and $(4,12)$ .
The $x$ coordinates of these points give the roots.

Answer

Solution set $\{-3,\ 4\}$

Q3. Draw the graph of $y = x^2 + x$ and hence solve $x^2 + 1 = 0$

Show steps (11)

Draw $y=x^2+x$ using the table:
$x$ : $-3,\ -2,\ -1,\ 0,\ 1,\ 2,\ 3$
$y$ : $6,\ 2,\ 0,\ 0,\ 2,\ 12$ → correctly $6,\ 2,\ 0,\ 0,\ 2,\ 6,\ 12$
To solve $x^2+1=0$ , subtract it from $y=x^2+x$ :
$y=x^2+x$
$0=x^2+1\quad(-)$
$y=x-1$
Table for the line $y=x-1$ :
$x$ : $-2,\ -1,\ 0,\ 1,\ 2$
$y$ : $-3,\ -2,\ -1,\ 0,\ 1$
The line $y=x-1$ neither cuts nor touches the parabola $y=x^2+x$ .

Answer

The line does not meet the parabola, so $x^2+1=0$ has no real roots

Q4. Draw the graph of $y = x^2 + 3x + 2$ and use it to solve $x^2 + 2x + 1 = 0$

Show steps (12)

Draw $y=x^2+3x+2$ using the table:
$x$ : $-4,\ -3,\ -2,\ -1,\ 0,\ 1,\ 2$
$y$ : $6,\ 2,\ 0,\ 0,\ 2,\ 6,\ 12$
To solve $x^2+2x+1=0$ , subtract it from $y=x^2+3x+2$ :
$y=x^2+3x+2$
$0=x^2+2x+1\quad(-)$
$y=x+1$
Table for the line $y=x+1$ :
$x$ : $-4,\ -2,\ -1,\ 0,\ 2$
$y$ : $-3,\ -1,\ 0,\ 1,\ 3$
The line touches the parabola at only one point $(-1,0)$ .
So $x=-1$ is the only solution.

Answer

$x=-1$ (real and equal roots)

Q5. Draw the graph of $y = x^2 + 3x - 4$ and hence use it to solve $x^2 + 3x - 4 = 0$

Show steps (8)

Draw $y=x^2+3x-4$ using the table:
$x$ : $-5,\ -4,\ -3,\ -2,\ -1,\ 0,\ 1,\ 2$
$y$ : $6,\ 0,\ -4,\ -6,\ -6,\ -4,\ 0,\ 6$
To solve $x^2+3x-4=0$ , subtract it from $y=x^2+3x-4$ :
$y=x^2+3x-4$
$0=x^2+3x-4\quad(-)$
$y=0$ , which is the $X$ axis.
The curve cuts the $X$ axis at $(-4,0)$ and $(1,0)$ .

Answer

Solution set $\{-4,\ 1\}$

Q6. Draw the graph of $y = x^2 - 5x - 6$ and hence solve $x^2 - 5x - 14 = 0$

Show steps (8)

Draw $y=x^2-5x-6$ using the table:
$x$ : $-2,\ -1,\ 0,\ 1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7$
$y$ : $8,\ 0,\ -6,\ -10,\ -12,\ -12,\ -10,\ -6,\ 0,\ 8$
To solve $x^2-5x-14=0$ , subtract it from $y=x^2-5x-6$ :
$y=x^2-5x-6$
$0=x^2-5x-14\quad(-)$
$y=8$ , a line parallel to the $X$ axis.
This line cuts the parabola at $(-2,8)$ and $(7,8)$ .

Answer

Solution set $\{-2,\ 7\}$

Q7. Draw the graph of $y = 2x^2 - 3x - 5$ and hence solve $2x^2 - 4x - 6 = 0$

Show steps (11)

Draw $y=2x^2-3x-5$ using the table:
$x$ : $-2,\ -1,\ 0,\ 1,\ 2,\ 3,\ 4$
$y$ : $9,\ 0,\ -5,\ -6,\ -3,\ 4,\ 15$
To solve $2x^2-4x-6=0$ , subtract it from $y=2x^2-3x-5$ :
$y=2x^2-3x-5$
$0=2x^2-4x-6\quad(-)$
$y=x+1$
Table for the line $y=x+1$ :
$x$ : $-2,\ -1,\ 0,\ 1,\ 3$
$y$ : $-1,\ 0,\ 1,\ 2,\ 4$
The line cuts the parabola at $(-1,0)$ and $(3,4)$ .

Answer

Solution set $\{-1,\ 3\}$

Q8. Draw the graph of $y = (x - 1)(x + 3)$ and hence solve $x^2 - x - 6 = 0$

Show steps (12)

$y=(x-1)(x+3)=x^2+2x-3$ .
Table of values:
$x$ : $-4,\ -3,\ -2,\ -1,\ 0,\ 1,\ 2,\ 3$
$y$ : $5,\ 0,\ -3,\ -4,\ -3,\ 0,\ 5,\ 12$
To solve $x^2-x-6=0$ , subtract it from $y=x^2+2x-3$ :
$y=x^2+2x-3$
$0=x^2-x-6\quad(-)$
$y=3x+3$
Table for the line $y=3x+3$ :
$x$ : $-2,\ -1,\ 0,\ 1,\ 3$
$y$ : $-3,\ 0,\ 3,\ 6,\ 12$
The line cuts the parabola at $(-2,-3)$ and $(3,12)$ .

Answer

Solution set $\{-2,\ 3\}$

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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Q1. Graph the following quadratic equations and state their nature of solutions.

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Q2. Draw the graph of y=x2−4y = x^2 - 4y=x2−4 and hence solve x2−x−12=0x^2 - x - 12 = 0x2−x−12=0

Q3. Draw the graph of y=x2+xy = x^2 + xy=x2+x and hence solve x2+1=0x^2 + 1 = 0x2+1=0

Q4. Draw the graph of y=x2+3x+2y = x^2 + 3x + 2y=x2+3x+2 and use it to solve x2+2x+1=0x^2 + 2x + 1 = 0x2+2x+1=0

Q5. Draw the graph of y=x2+3x−4y = x^2 + 3x - 4y=x2+3x−4 and hence use it to solve x2+3x−4=0x^2 + 3x - 4 = 0x2+3x−4=0

Q6. Draw the graph of y=x2−5x−6y = x^2 - 5x - 6y=x2−5x−6 and hence solve x2−5x−14=0x^2 - 5x - 14 = 0x2−5x−14=0

Q7. Draw the graph of y=2x2−3x−5y = 2x^2 - 3x - 5y=2x2−3x−5 and hence solve 2x2−4x−6=02x^2 - 4x - 6 = 02x2−4x−6=0

Q8. Draw the graph of y=(x−1)(x+3)y = (x - 1)(x + 3)y=(x−1)(x+3) and hence solve x2−x−6=0x^2 - x - 6 = 0x2−x−6=0

Quick check

Q2. Draw the graph of $y = x^2 - 4$ and hence solve $x^2 - x - 12 = 0$

Q3. Draw the graph of $y = x^2 + x$ and hence solve $x^2 + 1 = 0$

Q4. Draw the graph of $y = x^2 + 3x + 2$ and use it to solve $x^2 + 2x + 1 = 0$

Q5. Draw the graph of $y = x^2 + 3x - 4$ and hence use it to solve $x^2 + 3x - 4 = 0$

Q6. Draw the graph of $y = x^2 - 5x - 6$ and hence solve $x^2 - 5x - 14 = 0$

Q7. Draw the graph of $y = 2x^2 - 3x - 5$ and hence solve $2x^2 - 4x - 6 = 0$

Q8. Draw the graph of $y = (x - 1)(x + 3)$ and hence solve $x^2 - x - 6 = 0$