Question

find a quadratic polynomial whose zeroes are 2+√2 and 2-√2​

Answer 1

$\leadsto \sf \pink{ Solution:-}$

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$\pink \leadsto \footnotesize\sf{\alpha = 2 + \sqrt{2} \: \: \: \: and \: \: \: \: \beta = 2 - \sqrt{2} }$

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$\leadsto\footnotesize \sf \pink{We \: know \: that \: Sum \: of \: zeros = \frac{-b}{a}}$

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$\pink \leadsto \footnotesize\sf{\alpha + \beta = \frac{ - b}{a} }$

$\pink \leadsto \footnotesize\sf{ 2 + \sqrt{2} + 2 - \sqrt{2} = \frac{ - b}{a} }$

$\pink \leadsto \footnotesize\sf{ 2 + \cancel{\sqrt{2}} + 2 - \cancel{\sqrt{2}} = \frac{ - b}{a} }$

$\pink \leadsto \footnotesize\sf{ 4 = \frac{ - b}{a} }$

$\small \boxed{ \underline{ \pink \leadsto \footnotesize\sf{ - 4 = \frac{ b}{a} }}}$

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$\leadsto\footnotesize \sf \pink{ Now,Product \: of \: zeros = \frac{c}{a}}$

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$\pink \leadsto \footnotesize\sf{\alpha \beta = \frac{ c}{a} }$

$\pink \leadsto \footnotesize\sf{(2 + \sqrt{2}) \: ( 2 - \sqrt{2}) = \frac{c}{a} }$

$\pink \leadsto \footnotesize\sf{2 \: ( 2 - \sqrt{2}) + \sqrt{2} \: (2 - \sqrt{2}) = \frac{c}{a} }$

$\pink \leadsto \footnotesize\sf{4 - 2\sqrt{2}+ 2 \sqrt{2} - ( { \sqrt{2} })^{2} = \frac{c}{a} }$

$\pink \leadsto \footnotesize\sf{4 - \cancel{ 2\sqrt{2}}+ \cancel{2 \sqrt{2} } - 2 = \frac{c}{a} }$

$\pink \leadsto \footnotesize\sf{2 = \frac{c}{a} }$

$\small \boxed{ \underline{ \pink \leadsto \footnotesize\sf{2 = \frac{c}{a} }}}$

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$\leadsto\footnotesize \sf \pink{General \: Form \: of \: equation}$

$\pink\leadsto\footnotesize \sf{ {ax}^{2} + bx + c }$

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$\leadsto\footnotesize \sf \pink{From \: above \: the \: value \: of \: a \: is \: 1, b \: is \: -4 </strong><strong>\: </strong><strong>and \: c \: is \: 2}$

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$\pink\leadsto\footnotesize \sf{Now \: Substitute \: value \: of \: a,</strong><strong> </strong><strong>b \: and \: c}$

$\leadsto\footnotesize \sf \pink{ {ax}^{2} + bx + c }$

$\leadsto\footnotesize \sf \pink{ {x}^{2} - 4x + 2 }$

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$\footnotesize \sf { \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: OR}$

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$\leadsto\footnotesize \sf \pink{ {x}^{2} - sum \: of \: zeros(x) + product \: of \: zeros }$

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$\pink\leadsto\footnotesize \sf { Product \: of \: zeros = 2}$

$\pink\leadsto\footnotesize \sf{ Sum \: of \: zeros = 4}$

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$\leadsto\footnotesize \sf \pink{ {x}^{2} - sum \: of \: zeros(x) + product \: of \: zeros }$

$\leadsto\footnotesize \sf \pink{ {x}^{2} - 4x + 2 }$

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$\boxed{ \underline{\pink\leadsto\footnotesize \sf { Quadratic \: Equation \: is \: {x}^{2} - 4x + 2 }}}$

Answer 2

•Let the zeroes of the quadratic polynomial be α=2 ,β=22.

•Then, α+β=2 +22 =2 (1+2)=32.

•αβ=2 ×22 =4.

•Sum of zeroes =α+β=32.

•Then, the quadratic polynomial =x2−( sum of zeroes )x+ product of zeroes =x2−(32 )x+4.

•=x2−32 x+4.

hope it's helpful