Question

please solve this 9 question​

Answer 1

Answer:

${ \large \bold \green{ \underline{p = \frac{ - 4}{7} \: or \: p = 4}}}$

Step-by-step explanation:

$\mathfrak{ \large \bold \red{ \underline{ \underline{question}}}} \\ \implies{ \large{find \: the \: value \: of \: p :}} \\ { \large{(2p + 1) {x}^{2} - (7p + 2)x + (7p - 3) = 0}} \\ \\ { \large \green{ \underline{given : }}} \\ \implies{ \large{ \underline{equation \: has \: real \: and \: equal \: roots}}} \\ \\ { \large \bold \blue{ \underline{ \underline{solution : }}}} \\ \implies{ \large{a = (2p + 1) \: and \: b = - (7p + 2) \: and \: c = (7p - 3) }} \\ \implies{ \large \orange{discrimant : d = {b}^{2} - 4ac }} \\ \implies{ \large{( - (7p + 2)) {}^{2} - 4(2p + 1) \times (7p - 3)}} \\ \implies{ \large{(49 {p}^{2} + 28p + 4) - 4(14 {p}^{2} + p - 3) }} \\ \implies{ \large{49 {p}^{2} + 28p + 4 - 56 {p}^{2} - 4p + 12}} \\ \implies{ \large{ - 7 {p}^{2} + 24p + 16 }} \\ { \large \pink{since \: root \: are \: real \: and \: equal(given)}} \\ \implies{ \large{7 {p}^{2} - 12p - 16 = 0 }} \\ \implies{ \large{7 {p}^{2} - 28p + 4p - 16 = 0}} \\ \implies{ \large{7p(p - 4) + 4(p - 4) = 0}} \\ \implies{ \large{(7p + 4)(p - 4) = 0}} \\ \implies{ \large{either : (7p - 4) = 0 \: or \: (p - 4) = 0}} \\ \implies{ \large{p = \frac{ - 4}{7} \: or \: p = 4}} \\ \\ { \large \bold \green{ \underline{p = \frac{ - 4}{7} \: or \: p = 4}}}$