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प्रश्न
Graph the following quadratic equation and state their nature of solution
x2 – 4x + 4 = 0
उत्तर
Let y = x2 – 4x + 4
(i) Prepare the table of values for y = x2 – 4x + 4
| 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 | 36 | 25 | 16 | 9 | 4 | 1 | 0 | 1 | 4 |
(ii) Plot the points (– 3, 25) (– 2, 16) (– 1, 9) (0, 4) (1, – 1) (2, 0), (3, 1) and (4, 4)
(iii) Join the points by a free hand smooth curve.
(iv) The roots of the equation are the X-coordinates of the intersecting points of the curve with X-axis (2, 0) which is 2.
(v) Since there is only one point of intersection with the X-axis (2, 0).
∴ The solution set is 2.
The Quadratic equation x2 – 4x + 4 = 0 has real and equal roots.
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