Math, asked by tonnyjohnny2, 4 months ago

Find zeroes of polynomial p(x) = 4 √3 x ² + 5x -2√3 and verify relationship

Answers

Answered by BrainlyRish
4

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀☆ Given Polynomial : p(x) = 4 √3 x ² + 5x -2√3

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:\implies\sf 4\sqrt {3}x^2 + 5x - 2\sqrt {3} = 0 \\\\\\:\implies\sf 4\sqrt {3}x^2 + 8x-3x - 2\sqrt {3} = 0 \\\\\\:\implies\sf 4x(\sqrt {3}x + 2) -\sqrt{3} (\sqrt {3}x + 2) = 0 \\\\\\:\implies\sf (\sqrt {3}x + 2) (4x -\sqrt {3}) = 0 \\\\\\:\implies{\underline{\boxed{\frak{\pink{x = \dfrac{-2}{\sqrt {3}} \; \&\; \dfrac{\sqrt{3}}{\;4}}}}}}\;\bigstar \\\\\sf{\therefore\; Zeroes\; of \; the \: Given \; polynomial \; are \; \dfrac{-2}{\sqrt{3}} \:\&\; \dfrac{\sqrt {3}\: }{\; \: 4 \: }.}

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\bf{\dag} \: \: \underline{\textsf{Relation b/w Coefficients \& Zeroes \: :}}⠀⠀

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{\qquad\maltese\:\:\textsf{Sum of Zeroes :}} \\\\\dashrightarrow\sf\:\:\alpha +\beta= \dfrac{ - \:b \: \: \: }{ \: \: \: a \: \: \:}\\\\\\\dashrightarrow\sf \bigg(\dfrac{-2}{\sqrt{3}}\bigg) +\bigg(\dfrac{\sqrt{3}}{\;4}\bigg) = \dfrac{-5 \: \: }{ \: \: 4\sqrt{3}\: \: } \\\\\\\dashrightarrow{\underline{\boxed{\frak{ \dfrac{-5 \: \: }{ \: \: 4\sqrt{3}\: \: } = \dfrac{-5 \: \: }{ \: \: 4\sqrt{3}\: \: }}}}}

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{\qquad\maltese\:\:\textsf{Product of Zeroes :}}\\\\\dashrightarrow\sf\:\:\alpha\beta=\dfrac{c}{a}\\\\\\\dashrightarrow\sf \bigg(  \dfrac{-2}{\sqrt{3}} \bigg) \times \bigg(\dfrac{\sqrt{3}}{\;4}\bigg) = \dfrac{2\sqrt {3}\: \: }{ \: \: 4\sqrt{3}\: \: } \\\\\\\dashrightarrow{\underline{\boxed{\frak{   \dfrac{2\sqrt {3}\: \: }{ \: \: 4\sqrt{3}\: \: } =  \dfrac{2\sqrt {3}\: \: }{ \: \: 4\sqrt{3}\: \: }}}}}

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

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Answered by Anonymous
1

⠀⠀⠀☆ Given Polynomial : p(x) = 4 √3 x ² + 5x -2√3

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\begin{gathered}:\implies\sf 4\sqrt {3}x^2 + 5x - 2\sqrt {3} = 0 \\\\\\:\implies\sf 4\sqrt {3}x^2 + 8x-3x - 2\sqrt {3} = 0 \\\\\\:\implies\sf 4x(\sqrt {3}x + 2) -\sqrt{3} (\sqrt {3}x + 2) = 0 \\\\\\:\implies\sf (\sqrt {3}x + 2) (4x -\sqrt {3}) = 0 \\\\\\:\implies{\underline{\boxed{\frak{\pink{x = \dfrac{-2}{\sqrt {3}} \; \&\; \dfrac{\sqrt{3}}{\;4}}}}}}\;\bigstar \\\\\sf{\therefore\; Zeroes\; of \; the \: Given \; polynomial \; are \; \dfrac{-2}{\sqrt{3}} \:\&\; \dfrac{\sqrt {3}\: }{\; \: 4 \: }.}\end{gathered}

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