Question

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

Answer 1

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.

Answer 2

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:

${ \red{ \bf{ Let's\: substitute \:the\: given\: value}}}$

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

px² + 7x + q = 0

p(2/3)² + 7(2/3) + q = 0

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

${ \red{ \bf{ }}}$

By taking LCM

4p + 42 + 9q = 0

${ \red{ \bf{ 4p + 9q = – 42----------(1)}}}$

${ \red{ \bf{Now, }}}$

${ \red{ \bf{ substitute\: the\: value\: x = -3 }}}$

${ \red{ \bf{ in\: the \:expression,\: we\: get:}}}$

px² + 7x + q = 0

p(-3)² + 7(-3) + q = 0

9p + q – 21 = 0

9p + q = 21

${ \red{ \bf{ q = 21 – 9-----------(2)}}}$

${ \red{ \bf{ By \: substituting \: the\: value\: of \:q\: in\: eqn. (1),}}}$

${ \red{ \bf{ We\: get:}}}$

4p + 9q = – 42

4p + 9(21 – 9p) = -42

4p + 189 – 81p = -42

189 – 77p = -42

189 + 42 = 77p

231 = 77p

p = 231/77

${ \red{ \bf{ p = 3 }}}$

${ \red{ \bf{ Now, \:substitute\: the\: value\: of\: p\: in \:equation (2), }}}$

${ \red{ \bf{ We \: get: }}}$

q = 21 – 9p

= 21 – 9(3)

= 21 – 27

${ \red{ \bf{ = -6 }}}$

${ \red{ \bf{ ∴ Value\: of \:p \: is \: 3 \: and \: q \: is \: -6. }}}$