Saved Bookmarks
| 1. |
(kx)2 + kx + 1 = (-4x)2 - x |
| Answer» We have,{tex}kx^2 + kx + 1 = -4x^2 - x{/tex}{tex}\\Rightarrow{/tex}{tex}kx^2 + 4x^2 + kx + x + 1 = 0{/tex}{tex}\\Rightarrow{/tex}{tex} (k + 4)x^2 + (k + 1)x + 1 = 0{/tex}Here, a = k + 4, b = k + 1 and c = 1{tex}\\therefore{/tex}{tex}D = b^2 - 4ac{/tex}= (k + 1)2 - 4 {tex}\\times{/tex} (k + 4) {tex}\\times{/tex} (1){tex}= k^2 + 1 + 2k - 4k - 16{/tex}{tex}= k^2 - 2k - 15{/tex}{tex}\\Rightarrow{/tex}{tex}D = k^2 - 2k - 15{/tex}The given equation will have real and equal roots, ifD = 0{tex}\\Rightarrow{/tex}{tex}k^2 - 2k - 15 = 0{/tex}{tex}\\Rightarrow{/tex}{tex} k^2 - 5k + 3k - 15 = 0{/tex}{tex}\\Rightarrow{/tex} k(k - 5) + 3(k - 5) = 0{tex}\\Rightarrow{/tex} (k - 5)(k + 3) = 0{tex}\\Rightarrow{/tex} k - 5 = 0 or k + 3 = 0{tex}\\Rightarrow{/tex} k = 5 or k = -3 | |