Question

Find value of 'k' so that equation (k+4)x2+(k+1)x+1=0 has equal find value of k​

Answer 1

Answer:

HOW TO SOLVE?

$equal \: roots \: means \: discriminant \: is \: zero . \\ d = {b}^{2} - 4ac \\ so \: this \: means \: that \: {b}^{2} - 4ac = 0 \\ b = (k + 1) \\ a = (k + 4) \\ c = 1 \\ (k + 1) {}^{2} - 4(k + 4)(1) = 0 \\ {k}^{2} + 2k + 1 - 4k - 16 = 0 \\ {k}^{2} - 2k - 15 = 0 \\ split \: the \: middle \: term \: as \: - 5k \: and \: + 3k \\ {k}^{2} - 5k + 3k - 15 = 0 \\ k(k - 5) + 3(k - 5) = 0 \\ (k - 5)(k + 3) = 0 \\ so \: values \: of \: k \: are \: - 3 \: and \: 5 \: by \: \\ equating \: k - 5 \: and \: k + 3 \: to \: zero$

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