Question

ind the quadratic plynomial sum and product of whose zeroes are -1 and -20 respectievly.​

Answer 1

Step-by-step explanation:

step by step explanation is in the attachment

Answer 2

$\huge \sf \fbox\orange{A}\fbox\pink{n}\fbox\blue{s}\fbox\red{w}\fbox \purple{e}\fbox\red{r}$

• 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 :

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

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

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

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

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

$\begin{gathered}\qquad:\implies \sf x^2 - (sum\:of\:zeroes)x + Product \:of\:zeroes \:=\:0\\\end{gathered}$

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

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

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

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

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

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

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

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

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

Therefore,

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

Finding zeroes of Quadratic polynomial :

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

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

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

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

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

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

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

Therefore,

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

$\rule{300}{1.5}$

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

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