Question

if 2/3 and -3 are roots of the equation px²+7x+q=0 find p and q​

Answer 1

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

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

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

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

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

$\pink{}$$\red{Mark}$$\green{As}$$\blue{Brainliest}$$\orange{}$

Answer 2

Concept used-

Here the concept of zeros of Quadratic polynomial is used. The roots of the quadratic equation is given as 2/3 and -3. We have to subsitute the roots(zeros) separately in px²+7x+q=0. Then we get two equation, 4p+9q= -42 and 9p+q= 21. Solve them by substitution method. We will get the value of p and q.

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Solution

Roots

$\:\: \:\: \bull \: \frac{2}{3} \\ \\ \: \: \: \:\: \: \bull \: -3$

Quadratic equation

$\sf{px²+7x+q=0}$

$\mathtt{» \orange{First \: subsitute \: \frac{2}{3} }}$

$\mathtt{px²+7x+q}$

$\sf{\implies p×(\frac{2}{3}) ^{2}+7×\frac{2}{3} +q=0}$

$\sf{\implies \frac{4p}{9} +\frac{14}{3} +q = 0}$

$\mathtt{ » \purple{Take \:LCM}}$

$\sf{\implies \frac{4p+42+9q}{9} =0}$

$\sf{\implies 4p+9q= -42} \: \: \: \: ...(i)$

$\mathtt{» \orange{Now \: subsitute \: -3 }}$

$\mathtt{px²+7x+q}$

$\sf{\implies p×(-3)² +7×(-3) +q=0}$

$\sf{\implies 9p-21+q= 0}$

$\sf{\implies 9p+q= 21}. ...(ii)$

»Now solve the equations by substitution method

$\mathtt{»From \: equatîon \: (ii) \: we \: have \: q= 21-9p}$

»subsitute in equation (i)

$\sf{ \implies4p+9q= -42}$

$\sf{ \implies 4p+9(21-9p)= -42}$

$\sf{ \implies 4p + 189 -81 p = -42}$

$\sf{ \implies -77 p = -231}$

$\sf{ \implies p= \frac{231}{77}}$

$\sf{ \implies \green{ p= 3}}$

❖ therefore the value of p is 3.

» Now subsitute the value of p = 3 in any of the equation to get the value of q.

$\sf{ \implies q= 21-9p}$

$\sf{ \implies q= 21 - 9× 3}$

$\sf{ \implies q= 21- 27}$

$\sf{ \implies \red{q= -6}}$

❖ therefore the value of q is -6

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