Find a quadratic polynomial whose zeroes are 2+√2 and 2-√2

1.	Find a quadratic polynomial whose zeroes are 2+√2 and 2-√2
Answer» $\leadsto \sf \pink{ Solution:-}$ $\: \: \: \: \: \: \: \: \: \: \: \: \:$ $\pink \leadsto \footnotesize\sf{\alpha = 2 + \sqrt{2} \: \: \: \: and \: \: \: \: \beta = 2 - \sqrt{2} }$ $\: \: \: \: \: \: \: \: \: \: \: \: \:$ $\leadsto\footnotesize \sf \pink{We \: know \: that \: Sum \: of \: zeros = \frac{-b}{a}}$ $\: \: \: \: \: \: \: \: \: \: \: \: \:$ $\pink \leadsto \footnotesize\sf{\alpha + \beta = \frac{ - b}{a} }$ $\pink \leadsto \footnotesize\sf{ 2 + \sqrt{2} + 2 - \sqrt{2} = \frac{ - b}{a} }$ $\pink \leadsto \footnotesize\sf{ 2 + \cancel{\sqrt{2}} + 2 - \cancel{\sqrt{2}} = \frac{ - b}{a} }$ $\pink \leadsto \footnotesize\sf{ 4 = \frac{ - b}{a} }$ $\small \boxed{ \underline{ \pink \leadsto \footnotesize\sf{ - 4 = \frac{ b}{a} }}}$ $\: \: \: \: \: \: \: \: \: \: \: \: \:$ $\leadsto\footnotesize \sf \pink{ Now,Product \: of \: zeros = \frac{c}{a}}$ $\: \: \: \: \: \: \: \: \: \: \: \: \:$ $\pink \leadsto \footnotesize\sf{\alpha \beta = \frac{ c}{a} }$ $\pink \leadsto \footnotesize\sf{(2 + \sqrt{2}) \: ( 2 - \sqrt{2}) = \frac{c}{a} }$ $\pink \leadsto \footnotesize\sf{2 \: ( 2 - \sqrt{2}) + \sqrt{2} \: (2 - \sqrt{2}) = \frac{c}{a} }$ $\pink \leadsto \footnotesize\sf{4 - 2\sqrt{2}+ 2 \sqrt{2} - ( { \sqrt{2} })^{2} = \frac{c}{a} }$ $\pink \leadsto \footnotesize\sf{4 - \cancel{ 2\sqrt{2}}+ \cancel{2 \sqrt{2} } - 2 = \frac{c}{a} }$ $\pink \leadsto \footnotesize\sf{2 = \frac{c}{a} }$ $\small \boxed{ \underline{ \pink \leadsto \footnotesize\sf{2 = \frac{c}{a} }}}$ $\: \: \: \: \: \: \: \: \: \: \: \: \:$ $\leadsto\footnotesize \sf \pink{General \: Form \: of \: equation}$ $\pink\leadsto\footnotesize \sf{ {ax}^{2} + bx + c }$ $\: \: \: \: \: \: \: \: \: \: \: \: \:$ $\leadsto\footnotesize \sf \pink{From \: above \: the \: value \: of \: a \: is \: 1, b \: is \: -4 \: and \: c \: is \: 2}$ $\: \: \: \: \: \: \: \: \: \: \: \: \:$ $\pink\leadsto\footnotesize \sf{Now \: Substitute \: value \: of \: a, b \: and \: c}$ $\leadsto\footnotesize \sf \pink{ {ax}^{2} + bx + c }$ $\leadsto\footnotesize \sf \pink{ {x}^{2} - 4x + 2 }$ $\: \: \: \: \: \: \: \: \: \: \: \: \:$ $\footnotesize \sf { \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: OR}$ $\: \: \: \: \: \: \: \: \: \: \: \: \:$ $\leadsto\footnotesize \sf \pink{ {x}^{2} - sum \: of \: zeros(x) + product \: of \: zeros }$ $\: \: \: \: \: \: \: \: \: \: \: \: \:$ $\pink\leadsto\footnotesize \sf { Product \: of \: zeros = 2}$ $\pink\leadsto\footnotesize \sf{ Sum \: of \: zeros = 4}$ $\: \: \: \: \: \: \: \: \: \: \: \: \:$ $\leadsto\footnotesize \sf \pink{ {x}^{2} - sum \: of \: zeros(x) + product \: of \: zeros }$ $\leadsto\footnotesize \sf \pink{ {x}^{2} - 4x + 2 }$ $\: \: \: \: \: \: \: \: \: \: \: \: \:$ $\boxed{ \underline{\pink\leadsto\footnotesize \sf { Quadratic \: Equation \: is \: {x}^{2} - 4x + 2 }}}$

Answer»

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

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

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

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

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

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

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

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

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

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

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

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

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

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

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

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

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

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

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

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

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

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

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

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

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

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

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

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

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

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

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

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

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

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

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

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

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

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

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

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

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

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

Find a quadratic polynomial whose zeroes are 2+√2 and 2-√2

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

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Find a quadratic polynomial whose zeroes are 2+√2 and 2-√2