3x
[tex]3×2 – 4 \sqrt{3x \: + 4 = 0} [/tex]
-b2+- √4ac/2a​

2 thoughts on “3x<br />[tex]3×2 – 4 \sqrt{3x \: + 4 = 0} [/tex]<br />-b2+- √4ac/2a​”

1. Given:–

   • [tex]\sf{3x^2 – 4 \sqrt{3x + 4 = 0}}[/tex]

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   Solution:–

   According to the question,

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   [tex]\dashrightarrow\sf{x = \dfrac{-b\pm\sqrt{b^2 – 4ac}}{2a}}[/tex]

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   [tex]\dashrightarrow\sf{x = \dfrac{4\sqrt3\pm\sqrt{(-4\sqrt3)^2 – 4(3)(4)}}{2(3)}}[/tex]

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   [tex]\dashrightarrow\sf{x = \dfrac{4\sqrt3\pm\sqrt{48 – 48}}{6}}[/tex]

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   [tex]\dashrightarrow\sf{x = \dfrac{4\sqrt3\pm\,0}{6}}[/tex]

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   [tex]\dashrightarrow\sf{x = \dfrac{4\sqrt3}{6}}[/tex]

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   [tex]\dashrightarrow\sf{x = \dfrac{2\sqrt3}{3}}[/tex]

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   [tex]\dashrightarrow{\red{\bf{x=\dfrac{2}{\sqrt3}}}}[/tex]

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2. [tex]\textbf{Given:}[/tex]

   [tex]\mathsf{3x^2-4\sqrt{3}x+4=0}[/tex]

   [tex]\textbf{To find:}[/tex]

   [tex]\textsf{Solution of the given equation}[/tex]

   [tex]\textbf{Solution:}[/tex]

   [tex]\textsf{Consider,}[/tex]

   [tex]\mathsf{3x^2-4\sqrt{3}x+4=0}[/tex]

   [tex]\mathsf{Here,\;a=3,b=-4\sqrt3,c=4}[/tex]

   [tex]\mathsf{The\;roots\;of\;the\;given\;equation\;are}[/tex]

   [tex]\mathsf{x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}}[/tex]

   [tex]\mathsf{x=\dfrac{4\sqrt3\pm\sqrt{(-4\sqrt3)^2-4(3)(4)}}{2(3)}}[/tex]

   [tex]\mathsf{x=\dfrac{4\sqrt3\pm\sqrt{48-48}}{6}}[/tex]

   [tex]\mathsf{x=\dfrac{4\sqrt3\pm\,0}{6}}[/tex]

   [tex]\mathsf{x=\dfrac{4\sqrt3}{6}}[/tex]

   [tex]\mathsf{x=\dfrac{2\sqrt3}{3}}[/tex]

   [tex]\implies\boxed{\mathsf{x=\dfrac{2}{\sqrt3}}}[/tex]

   [tex]\textbf{Find more:}[/tex]

   a(x^2 + 1) = (a^2+ 1) x, a + 0 solve by formula method​

   https://brainly.in/question/16238292

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