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| 1. |
Find the value of k for which the equation x2+4x+k=0 has coincident roots |
| Answer» Ans. x2\xa0+ 4x + k = 0,\xa0On Comparing with ax2 + bx + c = 0,We get a = 1, b = 4 and c = k\xa0For Coincident roots D = 0=> b2 - 4ac = 0=> 42\xa0- 4 (1)(k) = 0=> 16 - 4k =0=> 4k = 16\xa0=> k = 4\xa0 | |