Question

find the root of equation √3x^2-8x+4√3.​

Answer 1

Step-by-step explanation:

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Answer 2

$\huge\bigstar\mathfrak\blue{\underline{\underline{SOLUTION}}}$

$let \: f(x) = \sqrt{3x} {}^{2} - 8x + 4 \sqrt{3} \\ \\ = > \sqrt{3} {x}^{2} - 8x + 4 \sqrt{3} = 0 \\ \\ = > \sqrt{3} {x}^{2} - 6x - 2x + 4 \sqrt{3} = 0 \\ \\ = > \sqrt{3x} (x - 2 \sqrt{3} ) - 2(x - 2 \sqrt{3} ) = 0 \\ \\ = > (x - 2 \sqrt{3} ) \: ( \sqrt{3x} - 2) = 0 \\ \\ = > x = 2 \sqrt{3} \: \: or \: \: x = \frac{2}{ \sqrt{3} } = \frac{2 \sqrt{3} }{3}$

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