Question

Find the zeros of the polynomial f(x) = x^2 -5 and verify the relationship between its zeros coefficients

please answer :)​

Answer 1

$\qquad \qquad \pmb{\mathbb{ POLYNOMIAL \:\::\:\: } \sf x^2 - 5 \:}$

⠀⠀⠀⠀⠀》 Finding out zeroes of Polynomial :

$\qquad \dashrightarrow \sf x^2 - 5 = 0 \:$

$\qquad \dashrightarrow \sf x^2 - (\sqrt{5})^2 = 0 \:$

As , We know that ,

$\qquad \dag\:\:\bigg\lgroup \sf{ Algebraic \:Indentity \:\:: a^2 - b^2 = ( a + b ) ( a - b ) }\bigg\rgroup$

$\qquad \dashrightarrow \sf x^2 - (\sqrt{5})^2 = 0\:$

$\qquad \dashrightarrow \sf \Big\{ x - \sqrt{5}\Big\} \Big\{ x + \sqrt{5})\Big\} = \:0 \:$

$\qquad \dashrightarrow \sf x = \:\pm\sqrt{5}\:$

$\qquad \dashrightarrow \sf x = \:\sqrt{5} \:\:or\:\:-\sqrt{5}\:$

$\dashrightarrow \underline{\boxed{\purple{\pmb{\frak{ \:x = \:\sqrt{5} \:\:or\:\:-\sqrt{5}\: \: }}}}}\:\:\bigstar$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

$\qquad \underline {\bf \:\: \maltese \:\:\purple { Verification\: \:\:between \:it's \:zeroes \:and \:coefficient's \:\:relationship \:\::\:}}\:$

$\qquad \qquad \pmb{\mathbb{ POLYNOMIAL \:\::\:\: } \sf x^2 - 5 \:}$

As, We know that ,

$\qquad\underline {\boxed {\pmb{ \:\maltese \;Sum \:of \:zeroes \:\:\red {\:( \:\alpha \: + \beta )}\::}}}$

$\qquad \dashrightarrow \sf \bigg( \alpha \:+ \beta \: \bigg) \:=\:\dfrac{\:-(Cofficient \:of \:x\:)\:\:}{Cofficient \:of \:x^2 \:}$

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

$\qquad \dashrightarrow \sf \bigg( \alpha +\:\beta \: \bigg) \:=\:\dfrac{-(\:Cofficient \:of \:x\:)}{Cofficient \:of \:x^2 \:}\\\\\qquad \dashrightarrow \sf \Big\{ \sqrt{5} \:+ (-\sqrt{5} )\: \Big\} \:=\:\dfrac{-(\:0\:)}{1 \:}\\\\\qquad \dashrightarrow \sf \Big\{ \sqrt{5} \:-\sqrt{5} \: \Big\} \:=\:0 \\\\\qquad \dashrightarrow \sf 0 \: \:=\:0\\\\ \dashrightarrow \underline{\boxed{\purple{\pmb{\frak{ \:0 \: \:=\:0\: }}}}}\:\:\bigstar$

⠀⠀⠀⠀⠀AND ,

$\qquad\underline {\boxed {\pmb{ \:\maltese \;Product \:of \:zeroes \:\:\red {\:( \:\alpha \: \beta )}\::}}}$

$\qquad \dashrightarrow \sf \bigg( \alpha \:\beta \: \bigg) \:=\:\dfrac{\:Constant \:Term\:\:}{Cofficient \:of \:x^2} \:$

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

$\qquad \dashrightarrow \sf \bigg( \alpha \:\beta \: \bigg) \:=\:\dfrac{\:Constant\:Term\:}{Cofficient \:of \:x^2 \:}\\\\\qquad \dashrightarrow \sf \Big\{ \sqrt{5} \: (-\sqrt{5} )\: \Big\} \:=\:\dfrac{\:5\:}{1 \:}\\\\\qquad \dashrightarrow \sf \Big\{ 5\: \Big\} \:=\:\cancel{\dfrac{\:5\:}{1 \:}}\\\\\qquad \dashrightarrow \sf 5 \:=\:5 \\\\ \dashrightarrow \underline{\boxed{\purple{\pmb{\frak{ \:5 \: \:=\:5\: }}}}}\:\:\bigstar$

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