Question

find zeroes of polynomial
$\sqrt{3x { }^{2} } - 8x + 4 \sqrt{3}$
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Answer 1

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Answer 2

Answer:

$\huge\mathfrak{ \pink{answer}}$

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

$= \sqrt{3} {x}^{2} - 6x - 2x + 4 \sqrt{3} = 0$

$= \sqrt{3} (x - \sqrt[2]{3}) - 2(x - \sqrt[2]{3} )$

So the required zeros are

$= x = \frac{2}{ \sqrt{3} } \: or \: x = \sqrt[2]{3}$