Question

$\bf\huge{\pink{Meow \:Meow ♡︎}}$

Find the zero of the following quadratic polynomial 3x²-5x-2 and verify the relationship between the zeros and the cofficent​

Answer 1

Given that , The Quadratic Polynomial is 3x²-5x-2 .

Exigency To Find : The Zeroes & Verify the relationship between the zeroes and the cofficient ?

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$\qquad \qquad \underline{\pmb{\mathbb{\bigstar \:\:QUADRATIC \:\:POLYNOMIAL \:\:}}}\::\:\underline {\sf 3x^2 - 5x - 2 \: \:}$

⠀⠀⠀》 Finding , the zeroes of Polynomial .

$\qquad:\implies\:\: \sf 3x^2\:\: -\:\: 5x\:\: - 2\:\: = \:0\:\:\\\\ \qquad:\implies \sf 3x^2\:\: - 6x \:\:+ x \:\:- 2 \:=\:0\\\\ \qquad:\implies \sf 3x\:\:( \:\:x -\:\: 2 \:\:) \:\: +\:\: 1\:\: (\:\: x \:\:- 2 \:\:) \:\:=\:\:0\\\\ \qquad:\implies \sf ( \:\:3x +\:\: 1 \:\:) \:\: (\:\: x \:\:- 2 \:\:) \:\:=\:0\:\\\\ \qquad:\implies \underline {\boxed {\pmb{\pink{ \frak { \: x\: \:\:=\:\:-\dfrac{1}{3}\:or\:2 \:\:}}}}}\:\:\bigstar$

$\qquad \therefore \underline {\sf Hence, \:The \:zeroes \:of \:given \:polynomial \:are \: \pmb{\bf -1/3 \:and \:2 \;}\:.}$

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$\qquad \bigstar \:\:\underline {\sf Verification\:of \:relationship \:b/w \: it's \:zeroes \:and \:coefficient's \:\::}\:\:$

⠀⠀⠀⠀⠀Here , The zeroes of Polynomial are -1/3 & 2 [ α & β ].

$\qquad \underline {\boxed {\pmb{ \:\maltese \:Sum \:\: of \:\:zeroes \:\:\purple{ \:(\: \alpha + \beta \;)\:} \:: \: }}}\\\\\dashrightarrow \sf \Big\{ \: \alpha \:+\:\beta \:\Big\} \:=\:\:\dfrac{ -( \:Coefficient \:of \:x\:)}{ Coefficient \:of \:x^2 } \: \\\\\dashrightarrow \sf \Bigg\{ \: 2 \:+\:\bigg( - \dfrac{1}{3}\bigg) \:\Bigg\} \:=\:\:\dfrac{ -( \:-5\:)}{ 3 } \: \\\\ \dashrightarrow \sf \Bigg( \: 2 \: - \dfrac{1}{3} \:\Bigg) \:=\:\:\dfrac{ 5\:}{ 3 } \: \\\\ \dashrightarrow \sf \Bigg( \: \dfrac{6 - 1 }{3} \:\Bigg) \:=\:\:\dfrac{ 5\:}{ 3 } \: \\\\\dashrightarrow \sf \: \dfrac{5 }{3} \: \:=\:\:\dfrac{ 5\:}{ 3 } \: \\\\ \dashrightarrow \underline {\boxed {\pmb{\pink{ \frak { \: \dfrac{5 }{3} \: \:=\:\:\dfrac{ 5\:}{ 3 } \:\:\:}}}}}\:\:\bigstar$

⠀⠀⠀⠀⠀AND ,

$\qquad \underline {\boxed {\pmb{ \:\maltese \:Product \:\: of \:\:zeroes \:\:\purple{ \:(\: \alpha \beta \;)\:} \:: \: }}}\\\\\dashrightarrow \sf \Big\{ \: \alpha \:\:\beta \:\Big\} \:=\:\:\dfrac{ \:Constant \:Term\:}{ Coefficient \:of \:x^2 } \: \\\\\dashrightarrow \sf \Bigg\{ \: 2 \:\times \:\bigg( - \dfrac{1}{3}\bigg) \:\Bigg\} \:=\:\:\dfrac{ \:-2\:}{ 3 } \: \\\\ \dashrightarrow \sf \Bigg( \: 2 \:\times \dfrac{-1}{3} \:\Bigg) \:=\:\:\dfrac{ -2\:}{ 3 } \: \\\\ \dashrightarrow \sf \Bigg( \: \dfrac{-2 }{3} \:\Bigg) \:=\:\:\dfrac{ -2\:}{ 3 } \: \\\\\dashrightarrow \sf \: \dfrac{-2 }{3} \: \:=\:\:\dfrac{ -2\:}{ 3 } \: \\\\ \dashrightarrow \underline {\boxed {\pmb{\pink{ \frak { \: -\dfrac{2 }{3} \: \:=\:\:-\dfrac{ 2\:}{ 3 } \:\:\:}}}}}\:\:\bigstar$

$\qquad \therefore \pmb{\bf Hence \: Verified \:!\:}$

#SéXyMama

Answer 2

Answer:

Refer the attachment for your answer.

Hope it helps..