Question

find a quadratic polynomial whose zeroes are given below: 2/3,-5/7​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

☯︎ Given that

• Zeroes of a quadratic polynomial are given as 2/3 and -5/7.

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$\sf\underline{Firstly, \:we'll \: find \: the \: sum \: of \: zeroes}$

$\boxed{\boxed{\sf\purple{\bigstar \: Sum \: of \: zeroes \:}}}$

$\sf\implies \: Sum \: of \: zeroes \: = \dfrac{2}{3} + ( - \dfrac{5}{7} )$

$\sf\implies \: Sum \: of \: zeroes \: = \dfrac{14 + ( - 15)}{21}$

$\sf\implies \: Sum \: of \: zeroes \: = \dfrac{14 - 15}{21}$

$\sf\implies \: Sum \: of \: zeroes \: = \boxed{\boxed{\sf\purple{ \dfrac{ - 1}{21} }}}\bigstar$

$\sf\underline{Now, \: we'll \: find \: the \: product \: of \: zeroes}$

$\boxed{\boxed{\sf\purple{\bigstar \: Product \: of \: zeroes \:}}}$

$\sf\implies \: product \: of \: zeroes \: = \dfrac{2}{3} \times (\dfrac{ - 5)}{7}$

$\sf\implies \: product \: of \: zeroes \: = \dfrac{2 \times ( - 5)}{21}$

$\sf\implies \: product \: of \: zeroes \: = \boxed{\boxed{\sf\purple{ \dfrac{ - 10}{21} }}} \bigstar$

$\sf\underline{As \: we \: know \: that : }$

$\boxed{\boxed{\sf\purple{quadratic \: polynomial \: = {x}^{2} - (sum \: of \: zeroes)x + product \: of \: zeroes}}}$

$\sf\underline{Put \: the \: values:}$

$\sf\implies \: quadratic \: polynomial \: = {x}^{2} - (\dfrac{ - 1}{21})x + (\dfrac{ - 10}{21} )$

$\sf\implies \: quadratic \: polynomial \: = {x}^{2} + \dfrac{x}{21} - \dfrac{10}{21}$

$\sf\underline{Multiply \: throughout \: by \: 21}$

$\sf\implies \: quadratic \: polynomial \: = 21( {x}^{2} + \dfrac{x}{21} - \dfrac{10}{21} )$

$\sf\implies \: quadratic \: polynomial \: = \boxed{\boxed{\sf\purple{21 {x}^{2} + x - 10 }}} \bigstar$

$\sf{Hence,}$

• $\sf\underline{The \: quadratic \: polynomial \: is \: \bold{ 21{x}^{2} + x - 10}}$