Question

Find the zeros of the quadratic polynomial under root 3x^2-8x+4 under root 3

Answer 1

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