(k+4)x² + (k+1) + 1 = 0 has real and equal roots
using the formulae b² - 4ac = 0
find the value of k
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(k+1)^2 - 4(k+4)1=0
k^2 + 2k + 1 - 4k - 16=0
k^2 - 2k -15=0
k^2 - 5k +3k - 15 =0
k(k-5) + 3(k-5) = 0
k=5 & k =-3
then anser is k= 5
i hope this will help u
k^2 + 2k + 1 - 4k - 16=0
k^2 - 2k -15=0
k^2 - 5k +3k - 15 =0
k(k-5) + 3(k-5) = 0
k=5 & k =-3
then anser is k= 5
i hope this will help u
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