Question

it's urgent please solve this question... i will give 5 star rating and mark brainliest if correct

Answer 1

Step-by-step explanation:

$for \: roots \: to \: be \: real \: \: \: and\: equal \: of \: a \: qudratic \: equation \\ \\ discriminant \: (d) \: = 0 \\ \\ d \: = {b}^{2} - 4ac \: = 0 \: \: \: \: \: \: .......(1) \\ \\ where \: a = 2k + 1 \\ b \: = 2(k + 3) \\ \: \: \: c = - (k + 5) \\ \\ putting \: the \: values \: of \: a \: \: b \: \: c \: in \: equation \: (1) \\ \\ {(2k + 6)}^{2} - 4 \times (2k + 1) \times ( - (k + 5)) = 0 \\ = > 4 {k}^{2} + 2 \times 2k \times 6 + {6}^{2} + 4(2 {k}^{2} + k + 10k + 5) = 0 \\ = > 4 {k}^{2} + 24k + 36 + 8 {k}^{2} + 4k + 40k + 20 = 0 \\ = > 12 {k}^{2} + 68k + 56 = 0 \\ = > 4(3 {k}^{2} + 17k + 14) = 0 \\ = > 3 {k}^{2} + 17k + 14 = 0 \\ = > 3 {k}^{2} + 14k + 3k + 14 = 0 \\ = > k(3k \: + 14) + 1(3k + 14) = 0 \\ = > (k + 1)(3k + 14) = 0 \\ = > k + 1 = 0 \: \: \: \: \: | \: 3k \: + 14 = 0 \\ = > \: \: k \: \: = \: \: - 1 \: \: \: | k \: \: = \frac{ - 14}{3}$