Formation of Quadratic Equations with Given Roots – SS1 Mathematics Lesson Note (First Term)

INTRODUCTION 

The following steps should be taken when using the graphical method to solve quadratic equations:

i. Use the given range of values of the independent variable (usually x ) to determine the corresponding values of the dependent variable (usually y ) by the quadratic equation or relation given. If the range of values of the independent variable is not given, choose a suitable one.

ii. From the results obtained in step (i), prepare a table of values for the given quadratic expression.

iii. Choose a suitable scale to draw your graph.

iv. Draw the axes and plot the points.

v. Use a broom or flexible curve to join the points to form a smooth curve.

 NOTE:

i. The roots of the equation are the points where the curve cuts the x-axis because along the x-axis y = 0

ii. The curve can be an inverted n – shaped parabola or it can be a v-shaped parabola.  It is an n-shaped parabola when the coefficient of x2 is negative and it is V- a shaped parabola when the coefficient of x2 is positive.  The maximum value of y occurs at the peak or highest point of the n-shaped parabola while the minimum value of y occurs at the lowest point of the V-shaped parabola.

iii. The curve of a quadratic equation is usually in one of three positions to the x-axis.

iv. The line of symmetry is the line which divides the curve of the quadratic equation into two equal parts.

Examples

1a.   Draw the graph of  y =11 + 8x – 2×2 from x = -2 to x = +6.

1. b) Hence find the approximate roots of the equation 2×2 – 8x – 11=0 
2. From the graph, find the maximum value of y.

2a.Given that y = 4×2 – 12x + 9 ,copy and complete the table below

1. Hence draw a graph and find the roots of the equation 4×2 – 12x + 9 = 0
2. c) From the graph, what is the minimum value of y?
3. d) From the graph, what is the line of symmetry of the curve?

Solutions:

1)  Y = 11 +8x -2×2

from x   =-2 to x = + 6

When x  =-2

Y=11+8(-2)-2(-2)2

Y = 11 – 16 -2 ( +4)

Y =11 -16 – 8

Y = -5 – 8 = -13.

When x = -1

Y= 11 + 8 (-1) -2  (-1)2

Y= 11 – 8 – 2 ( + 1)

Y = 11 – 8 -2

Y = 3 -2 = 1.

When x = 0

Y = 11 + 8 (0) – 2 (0) 2

Y = 11 + 0  – 2 x 0

Y =11+ 0 – 0

Y=11

When x=1

Y = 11 + 8 ( 1) -2 ( 1)2

Y = 11 + 8 – 2 x 1

Y = 19 -2 = 17

When x =2

Y = 11 + 8 (2) -2 (2)2

= 11 + 16 – 2 x 4

= 27 – 8 = 19

when x = 3

y = 11 + 8 ( 3) – 2 ( 3) 2

= 11 + 24 – 2 x 9

= 35 – 18 = 17

when x = 4

y = 11 + 8 (4) – 2 (4) 2

= 11 + 32 – 2 x 16

= 43 – 32 = 11

when x = 5

y = 11 + 8 (5) -2 ( 5)2

= 11 + 40 -2 x 25

= 51 – 50 = 1

when x = 6

y = 11 + 8 ( 6) – 2 (6)

= 11 + 48 -2 x 36

= 59 – 72

=-13

1. b) From the graph, the approximate roots of the equation are the points where the curve cuts the x-axis, this is so because 

   y = 11 + 8x – 2×2

-1 x y = -1 x (11)  + 8x ( – 1) – 2×2 (-1)

-y = -11  – 8x + 2×2

-y = 2×2 – 8x – 11 = 0

-1x – y = 0 x -1

i.e y = 0

Thus, from the graph, the roots of the equation 2×2 -8x – 11 = 0 are x = -1.1or x = 5.1

19. c) The maximum value of y = 19.

2a)

From the graph, the roots of the equation are the points where the curve touches the x-axis i.e x = 1.5 twice

From the graph, the minimum value of y = 0

 From the graph, the line of symmetry of the curve is line x = 1.5

ASSIGNMENT 

1. Prepare a table of values for the graph of y = x2 + 3x – 4 for values of x from – 6 to + 3
2. b) Use a scale of 1cm to 1 unit on both axes and draw the graph.
3. c) Find the least value of y
4. d) What are the roots of the equation x2 + 3x – 4 = 0?
5. e) Find the values of x when y = 1