Question

find the zeroes of the polynomial root 3 x square plus 4 root 3 minus 8 x​

Answer 1

Answer:

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