Question

root3x square-2x =8root3 find x​

Answer 1

Answer:

$\sqrt{3} {x}^{2} - 2x - 8 \sqrt{3} = 0$

$\sqrt{3} {x}^{2} - 4 \sqrt{3} x + 2 \sqrt{3} x - 8 \sqrt{3} = 0$

$\sqrt{3} x(x - 4) + 2 \sqrt{3} (x - 4) = 0$

$(x - 4)( \sqrt{3} + 2 \sqrt{3} ) = 0$

$x = 4 \: or \: x = \frac{-2 \sqrt{3} }{ \sqrt{3} } = - 2$

i hope it will help u