if 2/3 and -3 are the roots of the equation px² + x² + q =0 find the values of P and Q​

2 thoughts on “if 2/3 and -3 are the roots of the equation px² + x² + q =0 find the values of P and Q​”

1. Correct Question:

   if 2/3 and -3 are the roots of the equation px² + 7x + q =0 find the values of P and Q

   Solution:

   [tex] { \red{ \bf{ Let’s\: substitute \:the\: given\: value}}}[/tex]

   [tex] { \red{ \bf{ x = 2/3 \:in \:the \:expression,\: we \:get:}}}[/tex]

   px² + 7x + q = 0

   p(2/3)² + 7(2/3) + q = 0

   4p/9 + 14/3 + q = 0

   [tex] { \red{ \bf{ }}}[/tex]

   By taking LCM

   4p + 42 + 9q = 0

   [tex] { \red{ \bf{ 4p + 9q = – 42———-(1)}}}[/tex]

   [tex] { \red{ \bf{Now, }}}[/tex]

   [tex] { \red{ \bf{ substitute\: the\: value\: x = -3 }}}[/tex]

   [tex] { \red{ \bf{ in\: the \:expression,\: we\: get:}}}[/tex]

   px² + 7x + q = 0

   p(-3)² + 7(-3) + q = 0

   9p + q – 21 = 0

   9p + q = 21

   [tex] { \red{ \bf{ q = 21 – 9———–(2)}}}[/tex]

   [tex] { \red{ \bf{ By \: substituting \: the\: value\: of \:q\: in\: eqn. (1),}}}[/tex]

   [tex] { \red{ \bf{ We\: get:}}}[/tex]

   4p + 9q = – 42

   4p + 9(21 – 9p) = -42

   4p + 189 – 81p = -42

   189 – 77p = -42

   189 + 42 = 77p

   231 = 77p

   p = 231/77

   [tex] { \red{ \bf{ p = 3 }}}[/tex]

   [tex] { \red{ \bf{ Now, \:substitute\: the\: value\: of\: p\: in \:equation (2), }}}[/tex]

   [tex] { \red{ \bf{ We \: get: }}}[/tex]

   q = 21 – 9p

   = 21 – 9(3)

   = 21 – 27

   [tex] { \red{ \bf{ = -6 }}}[/tex]

   [tex] { \red{ \bf{ ∴ Value\: of \:p \: is \: 3 \: and \: q \: is \: -6. }}}[/tex]

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2. Answer:

   Nature of roots of Q.E can be understood by evaluating the discriminant of the given Q.E.

   Let’s find it.

   Δ=b2−4ac

   Δ=[5(p+q)2]−4(p−q)[−2(p−q)]

   =25(p2+q2+2pq)−8(p−q)2

   =25p2+25q2+50pq−8(p2+q2−2pq)

   =25p2+25q2+50pq−8p2−8q2+16pq

   =17p2+17q2+66pq

   If values of p & q are both positive, then the roots are real.

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