Question

Find the roots: 4√3 x2

+5x -2√3 = 0 ​

Answer 1

$\huge\underline\frak\purple{Solution}$

$\sf 4\sqrt{3}\;x^2+5x-2\sqrt{3}=0$

$\sf 4\sqrt{3}\;x^2+8x-3x-2\sqrt{3}=0$

$\sf 4x(\sqrt{3}\;x+2)-\sqrt{3}(\sqrt{3}\;x+2)=0$

$\sf (4x-\sqrt{3}\;)\;(\sqrt{3}\;x+2)=0$

Either

x= √3/4

or

x= -2/√3

Answer 2

${\bold {\huge {\underline {\mathfrak {\green{Answer}}}}}}$

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$\boxed{\red{\bold{Given:}}}$

$4\sqrt{3} {x}^{2} + 5x - 2\sqrt {3} = 0$

$\boxed{\red{\bold{Solution:}}}$

${\blue {\underline {Simply\: using \:Radical \:expression}}}$

$4\sqrt{3}x^2+ 5x - 2\sqrt {3} = 0$

$4 \sqrt{3}x^2 + 8x -3x - 2 \sqrt{3} = 0$

${\blue {\underline {Solve\: equation \:for \:x}}}$

$4x(\sqrt{3}x + 2) - \sqrt{3}(\sqrt{3}x + 2)$

$(\sqrt{3}x+2)(4x-\sqrt{3}) = 0$

${\blue {\underline {Find\: the \:values }}}$

$x = \displaystyle{\frac{-2}{\sqrt {3}}}$

$x =\displaystyle{\frac{\sqrt {3}}{4}}$

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