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| 1. |
(k+1)x*X- 2(k-1)x+1=0 |
| Answer» {tex}( k + 1 ) x ^ { 2 } + 2 ( k + 1 ) x + 1 = 0{/tex}has equal roots if\xa0D = 0i.e\xa0{tex}b ^ { 2 } = 4 a c{/tex}Here, a = (k+1), b = 2 (k+1), c = 1or,\xa0{tex}4 ( k + 1 ) ^ { 2 } = 4 ( k + 1 ){/tex}{tex}k ^ { 2 } + 2 k + 1 = k + 1{/tex}{tex}k^2+2k+1-k-1=0{/tex}{tex}k ^ { 2 } + k = 0{/tex}k(k + 1) = 0k = 0, - 1Since k = - 1 does not satisfy the equation{tex}\\Rightarrow{/tex}k=0 | |