Formation Of Quadratic Equations With Given Roots SS1 Mathematics Lesson Note
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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.
- b) Hence find the approximate roots of the equation 2×2 – 8x – 11=0
- From the graph, find the maximum value of y.
2a.Given that y = 4×2 – 12x + 9 ,copy and complete the table below
| x | -1 | 0 | 1 | 2 | 3 | 4 |
| 4×2 | 4 | 16 | 64 | |||
| -12x | 12 | -24 | -48 | |||
| +9 | 9 | 9 | 9 | |||
| y | 24 | 1 | 3 | 25 |
- Hence draw a graph and find the roots of the equation 4×2 – 12x + 9 = 0
- c) From the graph, what is the minimum value of y?
- 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
- 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
- c) The maximum value of y = 19.
2a)
| x | -1 | 0 | 1 | 2 | 3 | 4 |
| 4×2 | 4 | 0 | 4 | 16 | 36 | 64 |
| -12x | 12 | 0 | -12 | -24 | -36 | -48 |
| +9 | 9 | 9 | 9 | 9 | 9 | 9 |
| y | 24 | 9 | 1 | 1 | 9 | 25 |
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
- Prepare a table of values for the graph of y = x2 + 3x – 4 for values of x from – 6 to + 3
- b) Use a scale of 1cm to 1 unit on both axes and draw the graph.
- c) Find the least value of y
- d) What are the roots of the equation x2 + 3x – 4 = 0?
- e) Find the values of x when y = 1