Question

Find the zeroes of the following quadratic polynomial and verify the
relationship between the zeroes and the coefficients : 2x^2-3+5x.​

Answer 1

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀☆ GIVEN POLYNOMIAL : 2x² - 3 + 5x

$\qquad \dashrightarrow \sf 2x^2 - 3 + 5x \: = 0$

$\qquad \dashrightarrow \sf 2x^2 + 5x - 3 \: = 0$

$\qquad \dashrightarrow \sf 2x^2 - 6x - x - 3 \: = 0$

$\qquad \dashrightarrow \sf 2x ( x - 3 ) - 1 ( x + 3 ) \: = 0$

$\qquad \dashrightarrow \sf ( 2x - 1 ) ( x + 3 ) \: = 0$

$\qquad \dashrightarrow \underline{\pmb{\purple{\: x \:\:=\:\: -3\:\:or\:\: \dfrac{1}{2} \:\:}} }\:\:\bigstar$

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

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

⠀⠀⠀⠀⠀

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

⠀⠀⠀

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

⠀⠀⠀

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

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