Question

find the zeroes of the quadratic polynomial and verify the relationship between the zeroes and coefficient p(x)= x2 -2x -8
pls answer fast

Answer 1

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀☆ GIVEN POLYNOMIAL : x² - 2x - 8

$\qquad \dashrightarrow \sf x^2 - 2x - 8 \: = 0$

$\qquad \dashrightarrow \sf x^2 - 4x + 2x - 8 \: = 0$

$\qquad \dashrightarrow \sf x( x - 4) + 2 ( x - 4 )\: = 0$

$\qquad \dashrightarrow \sf ( x + 2 ) ( x - 4 )\: = 0$

$\qquad \dashrightarrow \sf x \: = \:- 2 \:\:or \: 4$

$\qquad \dashrightarrow \underline{\pmb{\purple{\: x \:\:=\:\: -2\:\:or\:\: 4 \:\:}} }\:\:\bigstar$

⠀⠀⠀⠀⠀$\therefore {\underline{ \sf \: Hence, \:\:The \:zeroes \:of\:Polynomial \:are\:\bf \:\: -2\:\:\sf and\:\bf\: 4 \: }}$

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$\qquad \bigstar \qquad \underline {\sf Relationship\: between \:zeroes \:of\ polynomial\:and \:the \:Cofficients\:}:$⠀⠀⠀

$\qquad\maltese\:\:\textsf{Sum of Zeroes :} \\\\\dashrightarrow\sf\:\:\alpha +\beta= \dfrac{ - \:( \:Cofficient \:of\:x\:)\: \: \: }{ \: \: \: Cofficient \:of\:x^2 \:\: \: \:}\\\\\\\dashrightarrow\sf \bigg(-2\bigg) + \bigg( 4\bigg) = \dfrac{-(-2)}{1}\\\\\\\dashrightarrow\sf \bigg(-2\bigg) + \bigg( 4\bigg) = 2 \\\\\\\dashrightarrow{\underline{\boxed{\frak{ 2 = 2 }}}}$

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$\qquad\maltese\:\:\textsf{Product of Zeroes :}\\\\\dashrightarrow\sf\:\:\alpha\beta=\dfrac{Constant\:Term}{Cofficient\:of\:x^2 \:}\\\\\\\dashrightarrow\sf \bigg(-2\bigg) \times \bigg( 4 \bigg) = \dfrac{-8 \: \: }{ \: \: 1 \: \: } \\\\\\\dashrightarrow{\underline{\boxed{\frak{- 8 = -8 }}}}$

⠀⠀⠀

$\qquad\quad\therefore{\underline{\pmb{\textbf{Hence, Verified!}}}}$

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