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प्रश्न
Draw the graph of y = x2 + 3x + 2 and use it to solve x2 + 2x + 1 = 0
उत्तर
(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
y = x2 + 3x + 2
0 = x2 + 2x + 1
(–) (–) (–)
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 intersects the curve at only one point (– 1, 0).
The solution set is (– 1).
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