Question

If p and q are the zeroes of 3x²-6x+4 , then value of p²+q² is


pls answer fast

Answer 1

$\large \bf \clubs \:  Given :-$

p and q are the zeroes of 3x² - 6x + 4

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$\large \bf \clubs \: To \: Find :-$

• Value of p² + q²

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$\large \bf \clubs \:  Main \: Concept :-$

✏ 1》For a qudratic Polynomial of the Form ax² + bx + c :-

• Sum of Zeros = $\sf-\dfrac{b}{a}$

• Product of Zeros =$\sf\dfrac{c}{a}$

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$\large \bf \clubs \:  Solution :-$

For Polynomial is 3x² - 6x + 4 :

$\text{Sum of Zeros} = - \bigg( \dfrac{ -6 }{3} \bigg)$

$\large :\longmapsto\bf p + q = 2 \: \: - - - (1)$

$\text{Product of Zeros} = \dfrac{4}{3}$

$\large:\longmapsto \bf pq = \dfrac{4}{3} \: \: - - - (2)$

$\Large\text{Now,}$

$\large \sf{p^2+ q^2}$

$\large = \sf{p^2 +q ^2+ 2pq} - \text{2pq}$

$\large= \sf{(p + q)^2} - 2 \text{pq}$

✏ Using (1) and (2)

$= {2}^{2} - 2 \times \dfrac{4}{3}$

$= 4 - \dfrac{8}{3}$

$= \dfrac{12 - 8}{3}$

$= \dfrac{4}{3}$

Hence,

$\Large \underline{ \pink{ \underline{ \pmb{ \frak{p^2+q^2 = \dfrac{4}{3} }}}}}$

$\Large\red{\mathfrak{  \text{W}hich \:\:is\:\: the\:\: required} }\\ \LARGE \red{\mathfrak{ \text{ A}nswer}}$

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Answer 2

$\huge { \red{ \boxed{ \green { \boxed{ \color{yellow}{ \boxed{ \color{silver}{ \boxed{ \pink { \boxed{ \color{skyblue}{ \boxed{ \color{maroon}{ \boxed{\bf \blue{ \frac{4}{3} }}}}}}}}}}}}} }}}$

Step-by-step explanation:

QUESTION :-

If p and q are the zeroes of 3x² - 6x + 4, then the value of p² + q² is

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SOLUTION :-

3x² - 6x + 4

In the standard form of quadratic equation, here

• a = 3
• b = - 6
• c = 4

We know that,

Sum of zeroes (p + q) = $\bf \dfrac{-b}{a}$

And,

Product of zeroes (pq) = $\bf \dfrac{c}{a}$

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So,

here

Sum of zeroes = $\bf \dfrac{-b}{a}$

$\\ \bf = > \frac{ - ( - 6)}{3} \\ \\ \bf = > \cancel{ \frac{6}{3} } {}^{2}$

So,

Sum of zeroes (p + q) = 2

And,

Product of zeroes = $\bf \dfrac{c}{a}$

$\\ \bf = > \frac{4}{3}$

So,

Product of zeroes = $\bf \dfrac{4}{3}$

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Now,

p² + q²

=> (p + q)² - 2pq ...[using identity a² + b² = (a + b)²-2ab]

[put the value of (p + q) and pq]

$\\ \bf = > {(2)}^{2} - 2 (\frac{4}{3} ) \\ \\ = > \bf4 - \frac{8}{3} \\ \\ \bf = > \frac{(4 \times 3) - 8}{3} \\ \\ \bf = > \frac{12 - 8}{3} \\ \\ \bf = > \red{ \frac{4}{3} }$

So,

p² + q² = $\bf \dfrac{4}{3}$

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Hope it helps.

#BeBrainly :-)