10th Maths - Book Back Answers - Chapter 3 Exercise 3.16 - English Medium Guides

 

SSLC / 10th - Maths - Book Back Answers - Chapter 3 Exercise 3.16 - English Medium

Tamil Nadu Board 10th Standard Maths - Chapter 3 Exercise 3.16: Book Back Answers and Solutions

    This post covers the book back answers and solutions for Chapter 3 Exercise 3.16 – Maths from the Tamil Nadu State Board 10th Standard textbook. These detailed answers have been carefully prepared by our expert teachers at KalviTips.com.

    We have explained each answer in a simple, easy-to-understand format, highlighting important points step by step under the relevant subtopics. Students are advised to read and memorize these subtopics thoroughly. Once you understand the main concepts, you’ll be able to connect other related points with real-life examples and confidently present them in your tests and exams.

    By going through this material, you’ll gain a strong understanding of Chapter 3 Exercise 3.16 along with the corresponding book back questions and answers (PDF format).

Question Types Covered:

• 1 Mark Questions: Choose the correct answer, Fill in the blanks, Identify the correct statement, Match the following 
• 2 Mark Questions: Answer briefly 
• 3, 4, and 5 Mark Questions: Answer in detail

All answers are presented in a clear and student-friendly manner, focusing on key points to help you score full marks.

All the best, Class 10 students! Prepare well and aim for top scores. Thank you!

Chapter 3 Algebra Ex 3.16

1. Graph the following quadratic equations and state their nature of solutions.
(i) x² – 9x + 20 = 0
(ii) x² – 4x + 4 = 0
(iii) x² + x + 7 = 0
(iv) x² – 9 = 0
(v) x² – 6x + 9 = 0
(vi) (2x – 3) (x + 2) = 0

 

(i) x² – 9x + 20 = 0
Answer Key:
Let y = x² – 9x + 20
(i) Prepare the table of values for y = x² – 9x + 20

x	-3	-2	-1	0	1	2	3	4	5	6
X2	9	4	1	0	1	4	9	16	25	36
-9x	27	18	9	0	-9	-18	-27	-36	-45	-54
20	20	20	20	20	20	20	20	20	20	20
y	56	42	30	20	12	6	2	0	0	2

(ii) Plot the points (-1, 30) (0,20) (1, 12) (2, 6) (3,2), (4, 0), (5, 0), (6,2) (omit the high value)
(iii) Join the points by a free hand smooth curve.
(iv) The roots of the equation are the X-coordinates of the intersecting points of the curve with X-axis (4, 0) and (5, 0)
There are two points of intersection with the X-axis at 4 and 5. The solution set is 4 and 5. The quadratic equation has real and unequal roots.
(v) Since there is two point of intersection with X-axis (different solution)
∴ The equation x² – 9x + 20 = 0 has real and unequal roots.

 (ii) x² – 4x + 4 = 0
Answer Key:

Let y = x² – 4x + 4
(i) Prepare the table of values for y = x² – 4x + 4

x	-4	-3	-2	-1	0	1	2	3	4
X2	16	9	4	1	0	1	4	9	9
-4x	16	12	8	4	0	-4	-8	-12	-16
4	4	4	4	4	4	4	4	4	4
y	36	25	16	9	4	1	0	1	4

(ii) Plot the points (-3,25) (-2,16) (-1, 9) (0,4) (1,-1) (2, 0), (3,1) and (4, 4)
(iii) Join the points by a free hand smooth curve.
(iv) The roots of the equation are the X-coordinates of the intersecting points of the curve with X-axis (2, 0) which is 2.
(v) Since there is only one point of intersection with X-axis (2, 0).
∴ The solution set is 2.
The Quadratic equation x² – Ax + 4 = 0 has real and equal roots.

(iii) x² + x + 7 = 0

Answer Key:
Let y = x² + x + 7
(i) Prepare the table of values for y = x² + x + 7

x	-4	-3	-2	-1	0	1	2	3	4
X2	16	9	4	1	0	1	4	9	16
x	-4	-3	-2	-1	0	1	2	3	4
7	7	7	7	7	7	7	7	7	7
y	19	13	9	7	7	9	13	19	27

(ii) Plot the points (-4,19) (-3,13) (-2, 9) (-1, 7) (0, 7) (1, 9), (2,13) (3,19) and (4,27)
(iii) Join the points by a free hand smooth curve.
(iv) The solution of the given quadratic equation are the X-coordinates of the intersecting points of the parabola with the X-axis.

Parabola-themed merchandise

(v) The curve does not intersecting the X-axis. There is no solution set.
The equation x² + x + 7 = 0 has no real roots.

 

(iv) x² – 9 = 0
Answer Key:
Let y = x² – 9
(i) Prepare the table of values for y = x² – 9

x	-4	-3	-2	-1	0	1	2	3	4
X2	16	9	4	1	0	1	4	9	16
-9	-9	-9	-9	-9	-9	-9	-9	-9	-9
y	7	0	-5	-8	-9	-8	-5	0	7

(ii) Plot the points (-4, 7) (-3, 0) (-2, -5) (-1, -8) (0, -9) (1, -8), (2, -5) (3, 0) (4, 7)
(iii) Join the points by a free hand smooth curve.
(iv) The curve intersect the X-axis at -3 and 3.
The solution is (-3, 3).
(v) Since there are two points of intersection -3 and 3 with the X-axis the quadratic equation has real and unequal roots.

 

(v) x² – 6x + 9 = 0
Answer Key:
Let y = x² – 6x + 9
(i) Prepare a table of values for y = x² – 6x + 9

x	-4	-3	-2	-1	0	1	2	3	4	5
X2	16	9	4	1	0	1	4	9	16	25
-6x	24	18	12	6	0	-6	-12	-18	-24	-30
9	9	9	9	9	9	9	9	9	9	9
y	49	36	25	16	9	4	1	0	1	4

(ii) Plot the points (-2,25) (-1,16) (0,9) (1,4) (2,1) (3,0), (4,1) and (5,4) on the graph using suitable scale (omit the points (-4, 49) and (-3, 36)
(iii) Join the points by a free hand smooth curve.
(iv) The X – coordinates of the point of intersection of the curve with X-axis are the roots of the , given equation, provided they intersect.
The solution is 3.
(v) Since there is only one point of intersection with X-axis the quadratic equation x² – 6x + 9 = 0 has real and equal roots.

(vi) (2x – 3) (x + 2) = 0
Answer Key:
y = (2x – 3) (x + 2)
= 2x² + 4x – 3x – 6
= 2x² + x – 6

(i) Prepare a table of values for y from x – 4 to 4

x	-4	-3	-2	-1	0	1	2	3	4
X2	16	9	4	1	0	1	4	9	16
2x2	32	18	8	2	0	2	8	18	32
x	-4	-3	-2	-1	0	1	2	3	4
-6	-6	-6	-6	-6	-6	-6	-6	-6	-6
y	22	9	0	-5	-6	-3	4	15	30

(ii) Plot the points (-4, 22) (-3, 9) (-2, 0) (-1, -5) (0, -6) (1, -3), (2, 4), (3, 15) and (4, 30).
(iii) Join the points by a free hand smooth curve.
(iv) The curve intersect the X – axis at (-2, 0) and (1$\frac{1}{2}$, 0)
∴ The solution set is (-2,1$\frac{1}{2}$)
(v) Since there are two points of intersection with X – axis, the quadratic equation has real and un – equal roots.

2. Draw the graph of y = x² – 4 and hence solve x² – x – 12 = 0
Answer Key:
(i) Draw the graph of y = x² – 4 by preparing the table of values as below.

x	-4	-3	-2	-1	0	1	2	3	4
X2	16	9	4	1	0	1	4	9	16
-4	-4	-4	-4	-4	-4	-4	-4	-4	-4
Y	12	5	0	-4	-3	0	4	5	12

(ii) Plot the points for the ordered pairs (-4, 12) (-3, 5) (-2, 0) (-1, -3) (0, -4) (1, -3), (2, 0), (3, 5) and (4, 12). Draw the curve with the suitable scale.
(iii) To solve x² – x – 12 = 0 subtract x² – x – 12 from y = x² – 4

$\frac{\underset{\underset{ (-) (+) (+)}{0=x^{2 }-x- 12}}{y=x^2-0-4}}{y= x+8}$

The equation y = x + 8 represents a straight line. Prepare a table for y = x + 8

x	-3	-1	0	2	4
y	5	7	8	10	12

(iv) Mark the point of intersection of the curve and the straight line is (-3, 5) and 4, 12)
∴ The solution set is (-3, 4) for x² – x – 12 = 0.

3. Draw the graph of y = x² + x and hence solve x² + 1 = 0
Answer Key:
Let y = x² + x
(i) Draw the graph of y = x² + x by preparing the table.

x	-4	-3	-2	-1	0	1	2	3	4
X2	16	9	4	1	0	1	4	9	16
Y=x2+x	12	6	2	0	0	2	6	12	20

(ii) Plot the points (-4, 12), (-3, 6), (-2, 2), (-1, 0), (0, 0), (1, 2), (2, 6), (3, 12) and (4, 20).
(iii) Join the points by a free hand to get smooth curve.
(iv) To solve x² + 1 = 0, subtract x² + 1 = 0 from x² + x we get.

$\frac{\underset{\underset{ (-)(-) (-)}{0=x^2 +0 +1}}{y=x^2+x+0}}{y= x-1 }$

The equation represent a straight line. Draw a line y = x – 1

x	-3	-1	0	2
y	-4	-2	-1	1

Observe the graph of y = x² + 1 does not interset the parabola y = x² + x.

Parabola-themed merchandise

This x² + 1 has no real roots.

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

Answer Key:  
(i) Draw the graph of y = x² + 3x + 2 preparing the table of values as below.

x	-4	-3	-2	-1	0	1	2	3	4
X2	16	9	4	1	0	1	4	9	16
3x	-12	-9	-6	-3	0	3	6	9	12
2	2	2	2	2	2	2	2	2	2
y	6	2	0	0	2	6	12	20	30

(ii) Plot the points (-4, 6), (-3, 2), (-2, 0), (-1, 0), (0, 2), (1, 6), (2, 12), (3, 20) (4, 30).
(iii) To solve x² + 2x + 1 = 0 subtract x² + 2x + 1 = 0 from y = x² + 3x + 2

$\frac{\underset{\underset{ (-)(-) (-)}{0=x^2 +2x +1}}{y=x^2+3x+2}}{y= x+1}$

(iv) Draw the graph of y = x + 1 from the table

x	-4	-2	-1	0	2	4
y	-3	-1	0	1	3	5

The equation y = x + 1 represent a straight line.
This line intersect the curve at only one point (-1, 0). The solution set is (-1).

Algebra worksheets

 

5. Draw the graph of y = x² + 3x – 4 and hence use it to solve x² + 3x – 4 = 0
Answer Key:
Let y = x² + 3x – 4
(i) Draw the graph of y = x² + 3x – 4

x	-5	-4	-3	-2	-1	0	1	2	3	4
X2	25	16	9	4	1	0	1	4	9	16
-3x	-15	12	-9	-6	-3	0	3	6	9	12
-4	-4	-4	-4	-4	-4	-4	-4	-4	-4	-4
y	6	0	-4	-6	-6	-4	0	-12	-10	24

(ii) Plot the points (-5, 6), (-4, 0), (-3, -4), (-2, -6), (-1, -6), (0, -4), (1, 0), (2, 6), (3, 14) on the graph using suitable scale.
(iii) Join the points by a free hand smooth curve.
The smooth curve is the graph of y = x² – 4x + 4
(iv) To solve x + 3x – 4 = 0, subtract x² + 3x – 4 = 0 from y = x² + 3x – 4.
y = 0
∴ The point of intersection with the x – axis is the solution set.
The solution set is -4 and 1.

6. Draw the graph of y = x² – 5x – 6 and hence solve x², – 5x – 14 = 0
Answer Key:

Let y = x² – 5x – 6
(i) Draw the graph of y = x² – 5x – 6 by preparing the table of values as below.

x	-4	-3	-2	-1	0	1	2	3	4	5	6	7	8
X2	16	9	4	1	0	1	4	9	16	25	36	49	64
-5x	20	15	10	5	0	-5	-10	-15	-20	-25	-30	-35	-40
-6	-6	-6	-6	-6	-6	-6	-6	-6	-6	-6	-6	-6	-6
y	30	18	8	0	-6	-10	-12	-12	-10	6	0	8	18

(ii) Plot the points (-3, 18), (-2, 8), (-1, 0), (0, -6), (1, -10), (2, -12), (3, -12), (4, -10), (5,-6), (6, 0) and (7, 8).
(iii) Join the points by a free hand to get smooth curve.
(iv) To solve x² – 5x – 14 = 0, subtract x² – 5x – 14 = 0 from y = x² – 5x – 6.

$\frac{\underset{\underset{ (-)(+) (+)}{0=x^2-5x-14}}{y=x^2-5x-6}}{y= 8}$

The equation y = 8 represent a straight line draw a straight line through y = 8 intersect the curve at two places. From the two points draw perpendicular line to the X – axis it will intersect at -2 and 7.
The solution is -2 and 7

 

7. Draw the graph of y = 2x² – 3x – 5 and hence solve 2x² – 4x – 6 = 0
Answer Key:
(i) Draw the graph of y = 2x² – 3x – 5 by preparing the table of values given below.

x	-4	-3	-2	-1	0	1	2	3	4
X2	16	9	4	1	0	1	4	9	16
2x2	32	18	8	2	0	2	8	18	32
-3x	12	9	6	3	0	-3	-6	-9	-12
-5	-5	-5	-5	-5	-5	-5	-5	-5	-5
y	39	22	9	0	-5	-6	-3	4	15

(ii) Plot the points (-3, 22), (-2, 9), (-1, 0), (0, -5), (1,-6), (2, -3), (3, 4), (4, 15) on the graph sheet using suitable scale.
(iii) To solve 2x² – 4x – 6 = 0 subtract 2x² – 4x – 6 = 0 from y = 2x² – 3x – 5

$\frac{\underset{\underset{ (-) (+) (+)}{0=2x^2 -4x -6}}{y=2x^2-3x-5}}{y= x+1}$

(iv) y = x + 1 represent a straight line.

x	-4	-2	0	2	3	4
y	-3	-1	1	3	4	5

The straight line intersect the curve at (-1, 0) and (3, 4). From the two point draw perpendicular lines to the X – axis it will intersect at -1 and 3.
The solution set is (-1, 3)

 

8. Draw the graph of y = (x – 1) (x + 3) and hence solve x² – x – 6 = 0
Answer Key:
y = (x – 1) (x + 3)
y = x² + 2x – 3
(i) Draw the graph of y = x² + 2x – 3 by preparing the table of values given below

x	-4	-3	-2	-1	0	1	2	3	4
X2	16	9	4	1	0	1	4	9	16
2x	-8	-6	-4	-2	0	2	4	5	8
-3	-3	-3	-3	-3	-3	-3	-3	-3	-3
y	5	0	-3	-4	-3	0	5	12	21

(ii) Plot the points (-4, 5), (-3, 0), (-2, -3), (-1, -4), (0, -3), (1, 0), (2, 5), (3, 12) and (4, 21) on the graph sheet using suitable scale.
(iii) To solve x² – x – 6 = 0 subtract x² – x – 6 = 0 from y = x2 + 2x – 3
(iv) Draw the graph of y = 3x + 3 by preparing the table.

x	-4	-2	0	2	3	4
y	-9	-3	3	9	12	15

(v) The straight line cuts the curve at (-2, -3) and (3, 12). Draw perpendicular lines from the point to X – axis.
The line cut the X – axis at -2 and 3.
The solution set is (-2, 3)