Question

Find the zeroes of the following polynomial and verify the relationship between the zeroes and their
coefficients. t²-1​

Answer 1

Answer:

Zeroes are 0 and.

Step-by-step explanation:

Given: Quadratic Polynomial , 25x² + 5x.

To find: Zeroes of Polynomial and Verify the relation between zeroes and coefficient.

first relation is sum of zeroes/roots =

LHS = α + β =

Step-by-step explanation:

Drop A Thank Pls

Answer 2

⠀⠀⠀⠀⠀⠀⠀⠀⠀☆⠀G I V E N ⠀ P O L Y N O M I A L : t² - 1

⠀⠀⠀》⠀Finding out zeroes of Polynomial :

$\qquad \dashrightarrow \sf t^2 \:\:\: -\:\:1 \:\:=\:\:0\:$

As , We know that ,

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

$\qquad \dashrightarrow \sf \:\:(\:t\;+\:1\:)\:(\:t\:-\:1\:) \:\:=\:\:0\:\\\\\qquad \dashrightarrow \sf x \:\:=\:\:1\:\:\:or\:\:x\:\:=\:\:-1\:\: \\\\\qquad \dashrightarrow \:\:\underline {\boxed{\purple {\pmb{\frak{\:\:x \:\:=\:\:\:\:1\:\:\;or\:\:\:-1\:\:}}}}}$

$\qquad \therefore \:\:\underline {\sf\:Hence \:,\:The \:zeroes \:of \:Polynomial \:\:are \:\:\bf 1 \:\sf and \: \bf \:-1\:\sf.}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀¤⠀Verifying relationship between zeroes and Cofficients :

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

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

⠀⠀⠀⠀⠀AND ,

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

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

⠀⠀⠀⠀⠀$\therefore {\underline {\pmb{\bf{ Hence, \:Verified \:}}}}$