Question

please solve this question......​

Answer 1

Answer:

${ \large \green{ \underline{answer : k = 2 \: or \: k = \frac{ - 10}{9} }}}$

Step-by-step explanation:

$\mathfrak{ \large \green{ \underline{ \underline{given}}}} \\ \implies{ \large{ - 4 \: is \: a \: root \: of \: the \: equation \: {x}^{2} + 2x + 4p = 0}} \\ \implies{ \large{substitute \: the \: value \: of \: x = - 4}} \\ \\ \mathfrak{ \large \red{ \underline{solution}}} \\ \implies{ \large{( - 4) {}^{2} + 2( - 4) + 4p = 0 }} \\ \implies{ \large{16 - 8 + 4p = 0}} \\ \implies{ \large{8 + 4p = 0}} \\ \implies{ \large{4p = - 8}} \\ \implies{ \large{p = - 2}} \\ { \large{ \underline{in \: quadratric \: equation \: is \: }}} \\ \implies{ \large{ {x}^{2} + p(x)(1 + 3k) + 7(3 + 2k) = 0 }} \\ { \large{put \: the \: value \: of \: p}} \\ \implies{ \large{ {x}^{2} - 2x(1 + 3k) + 7(3 + 2k) = 0}} \\ \mathfrak{ \large \blue{ \underline{where : }}} \\ \implies{ \large{a = 1 \: and \: b = - 2(1 + 3k) \: and \: c = 7(3 + 2k)}} \\ \\ \mathfrak{ \large{ \underline{find \: discrimant}}} \\ \implies{ \large{d = {b}^{2} - 4ac}} \\ \implies{ \large{ (- 2(1 + 3k)) {}^{2} - 4 \times 1 \times 7(3 + 2k) }} \\ \implies{ \large{4(1 + 9 {k}^{2} + 6k) - 28(3 + 2k) }} \\ \implies{ \large{36 {k}^{2} - 32k - 80 }} \\ \\ \mathfrak{ \large{ \underline{since \: root \: are \: real \: and \: equal}}} \\ \implies{ \large{d = 0}} \\ \implies{ \large{36 {k}^{2} - 32k - 80 = 0}} \\ \implies{ \large{9 {k}^{2} - 8k - 20 = 0 }} \\ \implies{ \large{9 {k}^{2} - 18k + 10k - 20 = 0 }} \\ \implies{ \large{9k(k - 2) + 10(k - 2) = 0}} \\ \implies{ \large{(k - 2)(9k + 10) = 0}} \\ \implies{ \large{k - 2 = 0 \: or \: 9k + 10 = 0}} \\ k = 2\: and \: k = \frac{ - 10}{9} \\ \\ { \large \pink{ \underline{answer : k = 2 \: or \: k = \frac{ - 10}{9} }}}$