The equation x2 + k{4x + K - 1 } + 2 has real and equal roots. Find the value of K ?

The equation x2 + k{4x + K - 1 } + 2 has real and equal roots. Find th

1.	The equation x2 + k{4x + K - 1 } + 2 has real and equal roots. Find the value of K ?
Answer» Given, x2 + k(4x + k - 1) + 2 = 0 x2 + 4kx + k2\xa0- k\xa0+ 2 = 0.Here, a =1, b = 4k, c = k2-k\xa0+ 2 = 0Now equation has real roots, D = 0b2\xa0- 4ac = 0(4k)2 - 4\xa0{tex}\\times{/tex} 1\xa0{tex}\\times{/tex} (k2-k+2) = 016k2 - 4k2\xa0+ 4k\xa0- 8 = 012k2 + 4k - 8 = 04(3k2 + k - 2) = 03k2 +3k - 2k - 2\xa0= 03k(k+1)-2(k+1) = 0(3k-2)(k+1) = 0k =\xa0{tex}\\frac 23 \\ and \\ -1{/tex}