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Questions:

Generic 8

15 x 8 = 120

Graph the following quadratic equations and state their nature of solutions.
x² + x + 7 = 0

Graph the following quadratic equations and state their nature of solutions.
x² - 4x + 4 = 0

Discuss the nature of solutions of the following quadratic equations.
x² + 2x + 5 = 0

Discuss the nature of solutions of the following quadratic equations.
x² - 8x + 16 = 0

Draw the graph of y = (x - 1) (x + 3) and hence solve x² - x - 6 = 0

Draw the graph of y = 2x² - 3x - 5 and hence solve 2x² - 4x - 6 = 0

Draw the graph of y = x² - 5x - 6 and hence solve x² - 5x - 14 = 0

Draw the graph of y = x² + 3x - 4 and hence use it to solve x² + 3x - 4 = 0

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

10.

Draw the graph of y = x² - 4 and hence solve x² + 1 = 0

11.

Draw the graph of y = x² - 4 and hence solve x² - x - 12 = 0

12.

Draw the graph of y = x² + x - 2 and hence solve x² + x - 2 = 0

13.

Draw the graph of y = x² + 4x + 3 and hence find the roots of x² + x + 1 = 0

14.

Draw the graph of y = 2x² and hence solve 2x² - x - 6 = 0

15.

Discuss the nature of solutions of the following quadratic equations.
x² + x - 12 = 0

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Answers Key:

Generic 8

15 x 8 = 120

x² + x + 7 = 0
Let y=x2+x+7
Step 1:

x	-4	-3	-2	-1	0	1	2	3	4
x²	16	9	4	1	0	1	4	9	16
7	7	7	7	7	7	7	7	7	7
y=x²-x+7	19	13	9	7	7	9	13	19	27

Step 2:
Points to be plotted: (-4, 19), (-3, 13), (-2, 9), (-1, 7), (0, 7), (1, 9), (2, 13), (3, 19), (4, 27)
Step 3:
Draw the parabola and mark the co-ordinates of the parabola which intersect with the x-axis.

Step 4:
The roots of the equation are the points of intersection of the parabola with the x axis. Here the parabola does not intersect the x axis at any point.
So, we conclude that there is no real roots for the given quadratic equation.

x² - 4x + 4 = 0

x	-4	-3	-2	-1	0	1	2	3	4
x²	16	9	4	1	0	1	4	9	16
-4x	16	12	8	4	0	-4	-8	-12	-16
4	4	4	4	4	4	4	4	4	4
y=x²-4x+4	36	25	16	9	4	1	0	1	4

Step 1: Points to be plotted: (-4,36), (-3, 25), (-2, 16), (-1, 9), (0, 4), (1, 1), (2, 0), (3, 1), (4,4)
Step 2: The point of intersection of the curve with x axis is (2, 0)
Step 3:

Since there is only one point of intersection with x axis, the quadratic equation X² - 4x + 4 = 0 has real and equal roots.
∴ Solution{2,2}

x² + 2x + 5 = 0
Let y = x² + 2x + 5
Step 1 Prepare a table of values for the equation y = x² + 2x + 5

x	-3	-2	-1	0	1	2	3
y	8	5	4	5	8	13	20

Step 2: Plot the above ordered pairs(x, y) on the graph using suitable scale.

Step 3: Join the points by a free-hand smooth curve this smooth curve is the graph of y = x² + 2x + 5
Step 4: The solutions of the given quadratic equation are the x coordinates of the intersecting points of the parabola the X axis.
Here the parabola doesn’t intersect or touch the X axis.
So, we conclude that there is no real root for the given quadratic equation.

x² - 8x + 16 = 0
Step 1 Prepare the table of values for the equation y = x² - 8x + 16

x	-1	0	1	2	3	4	5	6	7	8
y	25	16	9	4	1	0	1	4	9	16

Step 2: Plot the points for the above ordered pairs (x, y) on the graph using suitable scale.

Step 3: Draw the parabola and mark the coordinates of the parabola which intersect with the X axis.
Step 4: The roots of the equation are the x coordinates of the intersecting points of the parabola with the X axis (4,0) which is 4.
Since there is only one point of intersection with X axis, the quadratic equation x² - 8x + 16 = 0 has real and equal roots.

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	6	8
-3	-3	-3	-3	-3	-3	-3	-3	-3	-3
y=x²+2x-3	5	0	-3	-4	-3	0	5	12	21

Draw the parabola using the points (-4, 5), (-3, 0), (-2, -3), (-1, -4), (0, -3), (1, 0), (2, 5), (3,12), (4, 21)

is a straight line

x	-2	-1	0	2
3x	-6	-3	0	6
3	3	3	3	3
y=3x+3	-3	0	3	9

Plotting the points (-2, -3), (-1, 0), (0, 3), (2, 9), we get a straight line.
The points of intersection of the parabola with the straight line gives the roots of the equation. The coordinates of the points of intersection forms the solution set.
∴ Solution {-2, 3}

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=x²-3x-5	39	22	9	0	-5	-6	-3	4	15

Draw the parabola using the points (-4, 39), (-3, 22), (-2, 9), (-1, 10), (0, -5), (1, -6), (2, -3), (3, 4), (4, 15).

To solve 2x²- 4x - 6 = 0, subtract it from y = 2x²- 3x - 5
is a straight line

x	-2	0	2
1	1	1	1
y=x+1	-1	1	3

Draw a straight line using the points (-2, -1), (0, 1), (2, 3). The points of intersection of the parabola and the straight line forms the roots of the equation.
The x-coordinates of the points of intersection forms the solution set.
∴ Solution {-1, 3}

x	-5	-4	-3	-2	-1	0	1	2	3	4
x²	25	16	9	4	1	0	1	4	9	16
-5x	25	20	15	10	5	0	-5	-10	-15	-20
-6	-6	-6	-6	-6	-6	-6	-6	-6	-6	-6
y=x²+5x-6	44	30	18	8	0	-6	-10	-12	-12	-10

Draw the parabola using the points (-5, 44), (-4, 30), (-3, 18), (-2, 8), (-1, 10), (0, -6), (1, -10), (2, -12), (3, -12), (4, -10)

To solve the equation X²- 5x - 14 = 0, subtract X² - 5x - 14 = 0 from y = X² - 5x - 6.
is a straight line parallel to x axis.
The co-ordinates of the points of intersection of the line and the parabola forms the solution set for the
equation X² - 5x - 14 = 0.
∴ Solution {-2, 7}

y=x²+3x-4

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
-4	-4	-4	-4	-4	-4	-4	-4	-4	-4
y=x²+3x-4	0	-4	-6	-6	-4	0	6	14	24

Draw the parabola using the points (-4, 0), (-3, -4), (-2, -6), (-1, -6), (0, -4), (1, 0), (2, 6), (3, 14), (4,24).

To solve: X² + 3x - 4 = 0 subtract X² + 3x - 4 = 0 from y = X² + 3x - 4

The points of intersection of the parabola with the x axis are the points (-4, 0) and (1, 0), whose x - co-ordinates (-4, 1) is the solution, set for the equation X2 + 3x - 4 = 0.

-1

10.

x	-4	-3	-2	-1	1	2	3	4	5
x²	16	9	4	1	1	4	9	16	25
+x	-4	-3	-2	-1	1	2	3	4	5
y=x²+x	12	6	2	0	2	6	12	20	30

Draw the parabola by the plotting the points (-4, 12), (-3, 6), (-2, 2), (-1, 0), (0, 0), (1, 2), (2, 6), (3, 12), (4,20), (5, 30)

To solve: X²+ 1 = 0, subtract X² + 1 = 0 from y = X² + x.

This is a straight line.
Draw the line y = x - 1.

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

Plotting the points (-2, -3), (0, -1), (2, 1) we get a straight line. This line does not intersect the parabola. Therefore there is no real roots for the equation X² + 1 = 0.

11.

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
x²-4	12	5	0	-3	-4	-3	0	5	12

x	-4	-3	-2	-1	0	1	2	3	4
8	8	8	8	8	8	8	8	8	8
x-8	4	5	6	7	8	9	10	11	12

Point of intersection (-3,5), (4, 12) solution of x² -x - 12 = 0 is -3, 4

12.

Step 1 : Draw the graph of y = x² + x - 2 by preparing the table of values as below

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

Step 2 : To solve x² + x - 2 = 0 subtract x² + x - 2 = 0 from y = x² + x - 2

The equation y = 0 represents the X axis.
Step 3 : Mark the point of intersection of the curve x² + x - 2 with the X axis. That is (–2,0) and (1,0)
Step 4 : The x coordinates of the respective points form the solution set {−2,1} for x² + x - 2 = 0

13.

Step 1 : Draw the graph of y = x² + 4x + 3 by preparing the table of values as below

x	-4	-3	-2	-1	0	1	2
y	3	0	-1	0	3	8	15

Step 2 : To solve x² + x + 1 = 0, subtract x² + x + 1 = 0 from y = x² + 4x + 3 that is,

The equation represent a straight line. Draw the graph of y = 3x + 2 forming the table of values as below

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

Step 3 : Observe that the graph of y = 3x + 2 does not intersect or touch the graph of the parabola y = x² + 4x + 3.

Thus x² + x + 1 = 0 has no real roots.

14.

Step 1: Draw the graph of y = 2x² by preparing the table of values as below

x	-2	-1	0	1	2
y	3	2	0	2	8

Step 2 : To solve 2x² - x - 6 = 0, subtract 2x² - x - 6 = 0 from y = 2x²

The equation y = x + 6 represents a straight line. Draw the graph of y = x + 6 by forming table of values as below

x	-2	-1	0	1	2
y	4	5	6	7	8

Step 3 : Mark the points of intersection of the curve y = 2x² and the line y = x + 6. That is, (–1.5, 4.5) and (2,8)
Step 4 : The x coordinates of the respective points forms the solution set {–1.5,2} for 2x² - x - 6 = 0

15.

x² + x - 12 = 0
Step 1 Prepare the table of values for the equation y = x² + x - 12

x	-5	-4	-3	-2	-1	0	1	2	3	4
y	8	0	-6	-10	-12	-12	-10	-6	0	8

Step 2: Plot the points for the above ordered pairs (x, y) on the graph using suitable scale.

Step 3: Draw the parabola and mark the co-ordinates of the parabola which intersect the X axis.
Step 4: The roots of the equation are the x coordinates of the intersecting points (–4, 0) and (3,0)of the parabola with the X axis which are −4 and 3 respectively.
Since there are two points of intersection with the X axis, the quadratic equation x² + x - 12 = 0 has real and unequal roots