Question

$\underline{ \boxed{\mathbb{\red{question}}}} \\ \\ p \: and \: q are \: zeros \: of \: {3x}^{2} + 2 x - 9 \: then \: the \: value \: of \: \: P \: - Q \: \: is$
${ \boxed{\mathbb{\pink{ reward \: for \: correct \: answer}}}}$

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please give correct answer​

Answer 1

Step-by-step explanation:

required is 2x² - 5x + 3. polynomial 3x² - 5x + 2.

it is given that, p and q are zeroes of

sum of zeroes = coefficient of x/ coefficient of x²

or, p + q = -(-5)/3 = 5/3

product of zeroes = constant/coefficient

of x²

or, pq = 2/3

now we have to find the polynomial zeroes of which are 1/p and 1/q.now we have to find the polynomial zeroes of which are 1/p and 1/q.

sum of zeroes = 1/p + 1/q

= (p + q)/pq

from equations (1) and (2),

= (5/3)/(2/3) = 5/2

and products of zeros = 1/px 1/q = 1/pq

= 1/(2/3) = 3/2

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:p \: and \: q \: are \: zeroes \: of \: {3x}^{2} + 2x - 9$

We know that

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\bf\implies \: p + q = - \dfrac{2}{3}$

And,

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\bf\implies \:pq = - \dfrac{9}{3} = - 3$

Consider,

$\rm :\longmapsto\:p - q$

$\rm \:  =  \: \sqrt{ {(p - q)}^{2} }$

$\rm \:  =  \: \sqrt{ {(p + q)}^{2} - 4pq}$

$\: \: \: \: \: \: \: \red{[ \because \: {(x + y)}^{2} - {(x - y)}^{2} = 4xy ]}$

$\rm \:  =  \: \sqrt{ {\bigg[ - \dfrac{2}{3} \bigg]}^{2} - 4( - 3) }$

$\rm \:  =  \: \sqrt{\dfrac{4}{9} + 12 }$

$\rm \:  =  \: \sqrt{\dfrac{4 + 144}{9}}$

$\rm \:  =  \: \sqrt{\dfrac{148}{9}}$

$\rm \:  =  \: \sqrt{\dfrac{2 \times 2 \times 37}{3 \times 3}}$

$\rm \:  =  \: \dfrac{2}{3} \sqrt{37}$

Hence,

$\\ \purple{\rm\implies \:\boxed{\tt{ \: \: \: p - q = \frac{2}{3} \sqrt{37} \: \: \: \: }}}$

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More to know :-

$\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: a {x}^{3} + b {x}^{2} + cx + d, \: then}$

$\boxed{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{b}{a}}}$

$\boxed{ \bf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}}}$

$\boxed{ \bf{ \: \alpha \beta \gamma = - \dfrac{d}{a}}}$