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Given Polynomial : 6x² – 3 – 7x
Need to find: The Zeroes of the given polynomial & also we've to verify the relationship b/w it's zeroes & their coefficients.
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❭❭ Finding out the zeroes of Given polynomial :
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\begin{gathered}:\implies\sf 6x^2 - 3 - 7x \\\\\\:\implies\sf 6x^2 - 7x - 3 = 0 \\\\\\:\implies\sf 6x^2 + 2x - 9x - 3 = 0\\\\\\:\implies\sf 2x(3x + 1) -3(3x + 1)=0\\\\\\:\implies\sf (2x - 3)(3x + 1) = 0\\\\\\:\implies\underline{\boxed{\frak{\pmb{\blue{x = \dfrac{3}{2} \;or\;x = \dfrac{-1}{\;3}}}}}}\;\bigstar\end{gathered}:⟹6x2−3−7x:⟹6x2−7x−3=0:⟹6x2+2x−9x−3=0:⟹2x(3x+1)−3(3x+1)=0:⟹(2x−3)(3x+1)=0:⟹x=23orx=3−1x=23orx=3−1★
\therefore{\underline{\textsf{Hence,\;the\; zeroes\;of\;the\;given\; polynomial\;are\; \textbf{3/2 and -1/3}.}}}∴Hence,thezeroesofthegivenpolynomialare3/2 and -1/3.
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V E R I F I C A T I O N :
☆ If α and β are roots of any Quadratic equation (ax² + bx + c = 0) then Sum and Product is given by :
⠀⠀⠀⠀⋆ Sum (α + β) = (–b)/a
⠀⠀⠀⠀⋆ Product (α β) = c/a
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\begin{gathered}{\qquad\maltese\:\:\textsf{Sum of Zeroes :}} \\\\\twoheadrightarrow\sf \alpha + \beta = \dfrac{-b}{\;a}\\\\\\\twoheadrightarrow\sf \bigg\{\dfrac{3}{2} \bigg\}+ \bigg\{\dfrac{-1}{\;3} \bigg\} = \dfrac{-(-7)}{6}\\\\\\\twoheadrightarrow\sf \dfrac{7}{6} = \dfrac{-(-7)}{\;\;6} \\\\\\\twoheadrightarrow\underline{\boxed{\frak{\dfrac{7}{6} = \dfrac{7}{6}}}}\end{gathered}✠Sum of Zeroes :↠α+β=a−b↠{23}+{3−1}=6−(−7)↠67=6−(−7)↠67=67
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\begin{gathered}{\qquad\maltese\:\:\textsf{Product of Zeroes :}}\\\\\twoheadrightarrow\sf \alpha\; \beta = \dfrac{c}{a} \\\\\\\twoheadrightarrow\sf \bigg\{\dfrac{3}{2}\bigg\} \times\bigg\{ \dfrac{-1}{\;3} \bigg\}= \dfrac{-3}{\:6} \\\\\\\twoheadrightarrow\sf \cancel\dfrac{-3}{\:6} = \cancel\dfrac{-3}{ \: 6} \\\\\\\twoheadrightarrow\underline{\boxed{\frak{\dfrac{-1}{\;2} = \dfrac{-1}{\;2}}}}\end{gathered}✠Product of Zeroes :↠αβ=ac↠{23}×{3−1}=6−3↠6−3=6−3↠2−1=2−1
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\qquad\quad\therefore{\underline{\purple{\textsf{\textbf{Hence, Verified!}}}}}∴Hence, Verified!