Question

y one zero of the polynomial x² - 4x + 1 is
2 + √3 write the other Zero​

Answer 1

Appropriate Question :

• One zero of the polynomial x² - 4x + 1 is 2 + √3 . Write the other Zero .

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❍ Let's Consider $\alpha \:\:\& \:\:\beta \:\:$ be the two zeroes of Polynomial x² - 4x + 1 .

Given :

• Polynomial : x² - 4x + 1

There Cofficient :

• Cofficient of x² = 1
• Cofficient of x = -4
• Constant Term = 1

&

• $\alpha$ = 2 + √3

$\dag\:\:\it{ As,\:We\:know\:that\::}\\\bf Sum \:of\:Roots\::$

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

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

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

$\qquad :\implies \sf 2 + \sqrt {3} + \beta = \dfrac{-(Cofficient \:of\:x \: )}{Cofficient\:of\:x^2}$

$\qquad :\implies \sf 2 + \sqrt {3} + \beta = \dfrac{-(-4 \: )}{1}$

$\qquad :\implies \sf 2 + \sqrt {3} + \beta = \dfrac{4 \:}{1}$

$\qquad :\implies \sf 2 + \sqrt {3} + \beta = 4$

$\qquad :\implies \sf \beta = 4 - 2 - \sqrt {3}$

$\qquad :\implies \sf \beta = 2 - \sqrt {3}$

$\qquad \longmapsto \frak{\underline{\purple{\: \beta = 2 - \sqrt {3} }} }\bigstar$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:\beta\:\:or\:\:Other\:zero \:of\: Polynomial \:is\:\bf{ 2 - \sqrt {3} }}}}$

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