Question

find the quadratic polynomial whose sum and product of zeroes are root 2 and product is -3/2 also find its zeros​

Answer 1

Given : The Product of whose zeroes are $\sf \dfrac{-3}{2}$ and the sum of the zeros is $\sf\sqrt{2}$.

Exigency to find : The Quadratic Polynomial & it's zeroes

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀Finding Quadratic polynomial :

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

$\qquad \dag\:\:\bigg\lgroup \sf{ x^2 - (sum\:of\:zeroes)x + Product \:of\:zeroes \:=\:0\: }\bigg\rgroup$

⠀⠀⠀⠀⠀Here sum of zeroes are $\bf\sqrt{2}$  and the product of the zeros is $\bf\dfrac{-3}{2}$ .

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

$\qquad:\implies \bf Quadratic \:Polynomial \:\:: \sf x^2 - (sum\:of\:zeroes)x + Product \:of\:zeroes\: =\:0$

$\qquad:\implies \sf x^2 - (sum\:of\:zeroes)x + Product \:of\:zeroes \:=\:0$

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

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

$\qquad:\implies \sf x^2 - \sqrt{2}x - \dfrac{3}{2} \:0\;$

$\qquad:\implies \sf\dfrac{ 2x^2 - 2\sqrt{2}x + -3}{2} \:0\;$

$\qquad:\implies \sf 2x^2 - 2\sqrt{2}x + ( -3) \:=\:0\times 2\;$

$\qquad:\implies \sf 2x^2 - 2\sqrt{2}x -3 \:=\:0\;$

$\qquad:\implies \bf 2x^2 - 2\sqrt{2}x -3 \:=\:0\;$

$\qquad :\implies \frak{\underline{\purple{\: 2x^2 - 2\sqrt{2}x -3 \:\:\qquad \longrightarrow\:Required\:Quadratic\:Polynomial\:\:}} }\:\:\bigstar$

Therefore,

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

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀Finding zeroes of Quadratic polynomial :

⠀⠀⠀⠀⠀⠀⠀⠀⠀☆⠀P O L Y N O M I A L : $\sf 2x^2 - 2\sqrt{2}x -3 \:$

$\qquad:\implies \bf Polynomial \: \::\sf 2x^2 - 2\sqrt{2}x -3 \:=\:0\;$⠀⠀⠀⠀⠀

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

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

$\qquad:\implies \sf \sqrt {2}x (\sqrt{2} x + 1) - 3( \sqrt{2}x + 1 ) \:=\:0\;$⠀⠀⠀⠀

$\qquad:\implies \sf (\sqrt {2}x - 3) (\sqrt{2} x + 1) \:=\:0\;$⠀⠀⠀⠀⠀

$\qquad :\implies \frak{\underline{\purple{\:x \:= \:\: \dfrac{3}{\sqrt{2}} \:\:or\:\:\dfrac{-1}{\sqrt{2}} }} }\:\:\bigstar$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:The \: zeroes \:of\:Quadratic\:Polynomial\:\:\:are\:\bf{\:\dfrac{3}{\sqrt{2}} \:\:and\:\:\dfrac{-1}{\sqrt{2}} \: }}}}$

$\rule{300}{1.5}$

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

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