Question

$\bf\fbox\red{Question:-}$
Find a quadratic polynomial with zeroes as $\sqrt{2} , \frac{ - 3}{ \sqrt{2} } ?$
​

Answer 1

Given : The Zeroes of Polynomial are $\bf \sqrt{2} \:\:\&\:\: \dfrac{ - 3}{ \sqrt{2} }$

Exigency To Find : The Quadratic Polynomial .

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❍ Let's Consider the zeroes of the polynomial be $\alpha \:\:and \:\:\beta \:$ .

$\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\ \qquad\bigstar\:\:\bf Quadratic\:Polynomial\::$

$\qquad \dag\:\:\bigg\lgroup \sf{ Quadratic \:Polynomial \: =\:\: x^2 - ( \: \alpha \:+ \beta \:)x \: + ( \:\alpha \times \:\beta \:) \:\:=0\:}\bigg\rgroup$

⠀⠀⠀⠀⠀Here , $\: \alpha + \beta \:$ is the sum of Zeroes , $\: \alpha \times \beta \:$ is the product of Zeroes & $\alpha \:\:and \:\:\beta \:$ are zeroes of Polynomial.

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⠀⠀⠀⠀⠀Now ,

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$\qquad \maltese\:\:\textsf{Sum of Zeroes :} \\\\\dashrightarrow\sf\:\:\alpha +\beta \\\\\\\dashrightarrow\sf \bigg( \sqrt{2} \bigg) + \bigg(\dfrac{-3}{\sqrt{2}} \bigg) \\\\\\\dashrightarrow \sf \bigg( \dfrac{ 2 - 3 }{\sqrt{2}} \bigg)\\\\\\\dashrightarrow \sf \bigg( \dfrac{ -1 }{\sqrt{2}} \bigg) \\\\\\\dashrightarrow{\underline{\boxed{\frak{\:\:\alpha + \beta \:\:=\:\:\dfrac{-1}{\sqrt{2}} }}}}$

⠀⠀⠀AND ,

$\qquad \maltese\:\:\textsf{Product of Zeroes :}\\\\\dashrightarrow\sf\:\:\alpha\beta \\\\\\\dashrightarrow\sf \bigg(\sqrt {2} \bigg) \times \bigg(\dfrac{-3 \: \: }{ \: \: \sqrt{2} \: \: }\bigg) \\\\\\\dashrightarrow\sf \bigg(\dfrac{-3\: \times \:\sqrt{2} \: }{ \: \: \sqrt{2} \: \: }\bigg) \\\\\\\dashrightarrow\sf \bigg(\dfrac{-3 \times \cancel {\sqrt{2}} \: \: }{ \: \: \cancel{\sqrt{2} }\: \: }\bigg) \\\\\\\dashrightarrow{\underline{\boxed{\frak{\alpha \beta \:\:\:= \:\: -3 }}}}$

$\qquad \dashrightarrow \:\sf Quadratic \:Polynomial \: =\:\: x^2 - ( \: \alpha \:+ \beta \:) x\: + ( \:\alpha \times \:\beta \:) \:=\:0\:$

$\qquad \dashrightarrow \:\sf \:\: x^2 - ( \: \alpha \:+ \beta \:)x \: + ( \:\alpha \times \:\beta \:) \:=\:0\:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \dashrightarrow \:\sf \:\: x^2 - ( \: \alpha \:+ \beta \:)x \: + ( \:\alpha \times \:\beta \:) \:=\:0\:$⠀⠀

$\qquad \dashrightarrow \:\sf \:\: x^2 - \bigg( \: -\dfrac{1}{\sqrt{2}}\:\bigg)x \: + ( \:-3 \:) \:=\:0\:$⠀

$\qquad \dashrightarrow \:\sf \:\: x^2 - \bigg( \: -\dfrac{1}{\sqrt{2}}\:\bigg)x \: + ( \:-3 \:) \:=\:0\:$⠀

$\qquad \dashrightarrow \:\sf \:\: x^2 \: +\dfrac{1}{\sqrt{2}}\: x\: \:-3 \: \:=\:0\:$⠀

$\qquad \dashrightarrow \:\sf \:\: \dfrac{\sqrt{2}x^2 \:\: + \: 1 x\: - \: 3 \sqrt{2}\:}{\sqrt{2}}\: \: \: \:=\:0\:$⠀

$\qquad \dashrightarrow \:\sf \:\: \sqrt{2}x^2 \:\: + \: x\: - \: 3 \sqrt{2}\:\: \: \: \:=\:0\:\times \:\sqrt{2}$⠀

$\qquad \dashrightarrow \:\sf \:\: \sqrt{2}x^2 \:\: + \: x\: - \: 3 \sqrt{2}\:\: \: \: \:=\:0\:$⠀

$\qquad \dashrightarrow \pmb{\underline{\purple{\: Quadratic\:Polynomial\: =\:\sqrt{2}x^2 \:\: +\: x\: - \: 3 \sqrt{2}\:\: }} }\:\;\bigstar$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:Hence,\:The \:Quadratic \:Polynomial \:is\:\bf{ \:\sqrt{2}x^2 \:\: +\: x\: - \: 3 \sqrt{2}\:\: }}.}}$

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⠀⠀$\large {\boxed{\sf{\mid{\overline {\underline {\star More\:To\:know\::}}}\mid}}}$

$\qquad \qquad \boxed {\begin{array}{cc} \bf{\underline {\bigstar\:\: For \: a \:Quadratic \:Polynomial \::}}\\\\ \sf{ Whose \:\:zeroes \:\:are\:\:\alpha \:\&\;\: \beta\:\:} \\\\ 1)\:\: \alpha + \beta \: =\:\dfrac{-b}{a} \quad \bigg\lgroup \bf Sum\:of\;Zeroes \bigg\rgroup \\\\ 2)\:\: \alpha \times \beta \: =\:\dfrac{c}{a} \quad \bigg\lgroup \bf Product \:of\;Zeroes \bigg\rgroup \\\\ \end{array}}$

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Answer 2

$\huge\color{cyan}\boxed{\colorbox{black}{Answer :-}}$

Given : Two zeros of the polynomial

$\sqrt{2} \: and \: \frac{ - 3}{ \sqrt{2} }$

Exigency To Find : The quadratic polynomial with its two roots as √2 and -3/√2.

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⠀⠀⠀⠀⠀

$Let α= \sqrt{2} \: and \: β= \frac{ - 3}{ \sqrt{2} } \\ \\α+β= \sqrt{2} + \frac{ - 3}{ \sqrt{2} } \\ \\ = \frac{2 - 3}{ \sqrt{2} } \\ \\ = \frac{ - 1}{ \sqrt{2} } \\ \\ and \: \\ \\ α⋅β=( \sqrt{2} )×( \frac{ - 3}{ \sqrt{2} } ) \\ \\ = - 3$

$The \: required \: quadratic \: equation \: is \: \\ {x}^{2} −(α+β)x+α⋅β=0$

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$\begin{gathered}\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\end{gathered}$

$⇒ {x}^{2} −(− \frac{1}{ \sqrt{2} } )x+( - 3)=0 \\ \\ ⇒{x}^{2}+ \frac{1}{ \sqrt{2} } x - 3=0 \\ \\ ⇒ \sqrt{2} {x}^{2} + x - 3 \sqrt{2} =0$

★ Therefore, $\begin{gathered}\therefore {\underline{ \mathrm {\:Hence,\:the \: polynomial \: is \: \bf{ \sqrt{2}x ^{2} +x - 3 \sqrt{2} }}\:\: \:\:.}}\\\end{gathered}$