Question

find d zeroes of d quadratic polynomial √3x2 -8x +4√3

Answer 1

Given quadratic equation = √3x² - 8x + 4√3 = 0
We should factorize the equation first.
          =√3x²-6x-2x+4√3 = 0
          = √3x(x-2√3)-2(x-2√3) = 0 
          = (√3x-2) (x-2√3) = 0
          = (√3x-2) = 0 , (x-2√3) = 0
          = x = 2/√3 , x = 2√3.

Answer 2

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

$So\ the\ zeroes\ are:\ \frac{2}{ \sqrt{3} }\ and\ 2 \sqrt{3}.$