Question

find zeros of quadratic polynomial root 3 x square - 8 x + 4 root 3

Answer 1

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Answer 2

Hey herr is your answer
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given polynomial p(x) :

$\sqrt{3} {x}^{2} - 8x + 4 \sqrt{3}$
let p(x) = 0
then,
$\sqrt{3 } {x}^{2} - 8x + 4 \sqrt{3} = 0 \\ \\ \sqrt{3 } {x}^{2} - 6x - 2x + 4 \sqrt{3} = 0 \\ \\ \sqrt{3} x(x - 2 \sqrt{3} ) - 2(x - 2 \sqrt{3} ) = 0 \\ \\ ( \sqrt{3} x - 2)(x - 2 \sqrt{3} ) = 0 \\ \\ ( \sqrt{3} x - 2) = 0 \: \: and \: \: (x - 2 \sqrt{3} ) = 0 \\ \\ x = \frac{2}{ \sqrt{3} } \: \: and \: \: 2 \sqrt{3} \\ \\ x = \frac{2 \sqrt{3} }{3} \: \: and \: \: 2 \sqrt{3}$
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