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| 1. |
Find the value of k for which root are real and equal .K2x2 - 2(2k - 1)x + 1 |
| Answer» {tex}k^2x^2-2(2k-1)x+1{/tex}on comparing with standard quadratic equation, we get a = k2, b = 2(2k-1) and c = 1\xa0For real and equal roots. D = 0\xa0=> b2-4ac = 0\xa0{tex}[2(2k-1)]^2-4\\times k^2\\times 1= 0\\\\=> 4(4k^2+1-4k)-4k^2=0\\\\=> k^2+1-4k-k^2=0\\\\=> 1 = 4 k \\\\=> k = {1\\over 4}{/tex} | |