Find the zeroes of the following quadratic polynomial and verify the
relationship between the zeroes and the coefficients : 2

1 thought on “Find the zeroes of the following quadratic polynomial and verify the<br />relationship between the zeroes and the coefficients : 2”

1. ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀☆ GIVEN POLYNOMIAL : 2x² – 3 + 5x

   [tex]\qquad \dashrightarrow \sf 2x^2 – 3 + 5x \: = 0\\[/tex]

   [tex]\qquad \dashrightarrow \sf 2x^2 + 5x – 3 \: = 0\\[/tex]

   [tex]\qquad \dashrightarrow \sf 2x^2 – 6x – x – 3 \: = 0\\[/tex]

   [tex]\qquad \dashrightarrow \sf 2x ( x – 3 ) – 1 ( x + 3 ) \: = 0\\[/tex]

   [tex]\qquad \dashrightarrow \sf ( 2x – 1 ) ( x + 3 ) \: = 0\\[/tex]

   [tex]\qquad \dashrightarrow \underline{\pmb{\purple{\: x \:\:=\:\: -3\:\:or\:\: \dfrac{1}{2} \:\:}} }\:\:\bigstar \\[/tex]

   ⠀⠀⠀⠀⠀[tex]\therefore {\underline{ \sf \: Hence, \:\:The \:zeroes \:of\:Polynomial \:are\:\bf \:\: -3\:\:\sf and\:\bf\: \dfrac{1}{2} \: }}\\[/tex]

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   [tex]\qquad \bigstar \qquad \underline {\sf Relationship\: between \:zeroes \:of\ polynomial\:and \:the \:Cofficients\:}: \\[/tex]

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   [tex]\qquad\maltese\:\:\textsf{Sum of Zeroes :} \\\\\dashrightarrow\sf\:\:\alpha +\beta= \dfrac{ – \:( \:Cofficient \:of\:x\:)\: \: \: }{ \: \: \: Cofficient \:of\:x^2 \:\: \: \:}\\\\\\\dashrightarrow\sf \bigg(-3\bigg) + \bigg(\dfrac{1}{ \: \: 2}\bigg) = \dfrac{-5}{2} \\\\\\\dashrightarrow{\underline{\boxed{\frak{\dfrac{-5}{2} = \dfrac{-5}{2}}}}} [/tex]

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   [tex]\qquad\maltese\:\:\textsf{Product of Zeroes :}\\\\\dashrightarrow\sf\:\:\alpha\beta=\dfrac{Constant\:Term}{Cofficient\:of\:x^2 \:}\\\\\\\dashrightarrow\sf \bigg(-3\bigg) \times \bigg(\dfrac{1 \: \: }{ \: \: 2 \: \: }\bigg) = \dfrac{-3 \: \: }{ \: \: 2 \: \: } \\\\\\\dashrightarrow{\underline{\boxed{\frak{\dfrac{-3}{\;2} = \dfrac{-3}{\;2}}}}}[/tex]

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   [tex]\qquad\quad\therefore{\underline{\pmb{\textbf{Hence, Verified!}}}}[/tex]

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