Can you answer this ....
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as it have real root so D = 0
D = 0
b²- 4ac
=> 4( k +1)² - 4 ( k + 1 ) × 4 = 0
k²+ 1 + 2k -4k -4 = 0
k²-2k -3 = 0
k² -3k+ k -3 = 0
k ( k -3 ) + (k -3 ) = 0
k = 3, -1
hence the value of k = 3,-1
hope it help you !!
thanks !!
D = 0
b²- 4ac
=> 4( k +1)² - 4 ( k + 1 ) × 4 = 0
k²+ 1 + 2k -4k -4 = 0
k²-2k -3 = 0
k² -3k+ k -3 = 0
k ( k -3 ) + (k -3 ) = 0
k = 3, -1
hence the value of k = 3,-1
hope it help you !!
thanks !!
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