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| 1. |
For what value of K are the roots of the quadratic equation kx(x-2)+6=0\xa0 |
| Answer» For Equal roots D = 0kx(x-2) + 6 = 0kx2 -2kx + 6 = 0comparing with ax2+ bx + c = 0, we get\xa0a = k, b = -2k , c = 6D = b2 - 4acPut value of a b c(-2k)2 - 4*6*k = 04k2 - 24k = 04k(k-6) =0\xa04k = 0 or k-6=0k = 6 | |