Question

plzz answer this question

Answer 1

Hi...☺

Here is your answer...✌

Given polynomial :

√3 x² - 8x + 4√3

To find zeros of polynomial, we put

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

Hence,

Zeros of given polynomial are 2√3 and 2/√3