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| 1. |
Find the value of k for which the quadratic equation (k+4)xsquar+(k+1)x+1 |
| Answer» We have, (k+4) x2 + (k+1)x + 1= 0Here a = (k+4), b = (k+1), c =1{tex}\\implies{/tex}D = b2 -4ac = (k+1)2 - 4 (k+4) (1)= k2 +1 + 2k - 4k -16= k2 -2k -15For equal roots, D = 0{tex}\\implies{/tex}k2 - 2k -15 = 0{tex}\\implies{/tex}k2 - 5k + 3k -15 = 0{tex}\\implies{/tex}k (k - 5) +3 (k - 5) = 0{tex}\\implies{/tex}(k+3) (k-5) = 0Either k+3 = 0 or k-5 = 0{tex}\\implies{/tex}k = -3, 5 | |