I.Solve the following equations graphically.
(i) x2 – 4x = 0
(ii) x2 + x – 12 = 0
(iii) x2 – x – 2 = 0
(iv) x2 – 5x + 6 = 0
II.
- Draw the graph of y = x2 and find the value of √3
- Draw the graph of y = 2x2 and find the value of √7
- Draw the graph of y = 1/2y2 and find the value of √10
Quadratic Equations – Exercise 9.10 – Solutions:
I.Solve the following equations graphically.
(i) x2 – 4x = 0
Solution:
Prepare the table of the values for the equation y = x2 – 4x
| x | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
| y | 5 | 0 | -3 | -4 | -3 | 0 | 5 |
The parabola intersects the x – axis at (0, 0) and (4, 0)
Therefore, the roots of equation are 0 and 4.
(ii) x2 + x – 12 = 0
Solution:
Prepare the table of the values for the equation x2 + x – 12 = 0
| x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
| y | 0 | 6 | -10 | -12 | -12 | -10 | -6 | 0 | 8 |
The parabola intersects the x – axis only at (3, 0) and (-4, 0)
Therefore, the roots of equation are 3 and -4.
(iii) x2 – x – 2 = 0
Solution:
Prepare the table of the values for the equation x2 – x – 2 = 0
| x | -2 | -1 | 0 | 1 | 2 | 3 |
| y | 4 | 0 | -2 | -2 | 0 | 4 |
The parabola intersects the x – axis at (-1, 0) and (2, 0)
Therefore, the roots of equation are -1 and 2.
(iv) x2 – 5x + 6 = 0
Solution:
Prepare the table of the values for the equation y = x2 – 5x + 6
| x | 0 | 1 | 2 | 3 | 4 | 5 |
| y | 6 | 2 | 0 | 0 | 2 | 6 |
The parabola intersects the x – axis at (2, 0) and (3, 0)
Therefore, the roots of equation are 2 and 3.
II.
- Draw the graph of y = x2 and find the value of √3
Solution:
y = x2
| x | 0 | -1 | 1 | -2 | 2 | √3 |
| y | 0 | 1 | 1 | 4 | 4 | 3 |
When x = √3, y = (√3)2 = 3
Draw a straight line y = 3 parallel to x-axis
The point on x – axis at which the perpendiculars meets are the values of √3 , x = ±1.7
Therefore, √3 = ±1.7
- Draw the graph of y = 2x2 and find the value of √7
Solution:
y = 2x2
| x | 0 | -1 | 1 | -2 | 2 | -3 | 3 | √7 |
| y | 0 | 2 | 2 | 8 | 8 | 18 | 18 | 14 |
When x = √7, y = 2(√7)2 = 14
Draw a straight line y = 14 parallel to x-axis.
The point on x – axis at which the perpendiculars meets are the values of √7 , x = ±2.6
Therefore, √7 = ±2.6
- Draw the graph of y = 1/2x2 and find the value of √10
Solution:
y = 1/2x2
| x | 0 | -2 | 2 | -4 | 4 | √10 |
| y | 0 | 2 | 2 | 8 | 8 | 5 |
When x = √10, y = 1/2(√10)2 = 5
Draw a straight line y = 5 parallel to x-axis.
The point on x – axis at which the perpendiculars meets are the values of √10 , x = ±3.1
Therefore, √10 = ±3.1
