⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀☆ Given Polynomial : p(x) = 4 √3 x ² + 5x -2√3 ⠀⠀⠀ [tex]:\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 \: }.}[/tex] [tex]\rule{250px}{.3ex}[/tex] [tex]\bf{\dag} \: \: \underline{\textsf{Relation b/w Coefficients \& Zeroes \: :}}[/tex]⠀⠀ ⠀⠀⠀⠀⠀ [tex]{\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}\: \: }}}}}[/tex] ⠀⠀⠀ [tex]{\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}\: \: }}}}}[/tex] ⠀⠀⠀ [tex]\qquad\quad\therefore{\underline{\textsf{\textbf{Hence, Verified!}}}}[/tex] ⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀ Reply
⠀⠀⠀☆ Given Polynomial : p(x) = 4 √3 x ² + 5x -2√3 ⠀⠀⠀ [tex]\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} [/tex] Reply
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀☆ Given Polynomial : p(x) = 4 √3 x ² + 5x -2√3
⠀⠀⠀
[tex]:\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 \: }.}[/tex]
[tex]\rule{250px}{.3ex}[/tex]
[tex]\bf{\dag} \: \: \underline{\textsf{Relation b/w Coefficients \& Zeroes \: :}}[/tex]⠀⠀
⠀⠀⠀⠀⠀
[tex]{\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}\: \: }}}}}[/tex]
⠀⠀⠀
[tex]{\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}\: \: }}}}}[/tex]
⠀⠀⠀
[tex]\qquad\quad\therefore{\underline{\textsf{\textbf{Hence, Verified!}}}}[/tex]
⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀
⠀⠀⠀☆ Given Polynomial : p(x) = 4 √3 x ² + 5x -2√3
⠀⠀⠀
[tex]\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} [/tex]