Question

find the zeroes of the polynomial 4√3 x²+5x_2√3 

Answer 1

4√3 x²+5x_2√3=0 => 4√3x^2 + 8x-3x-2√3 => 4x(√3x+2)-√3(√3x+2) => (4x-√3)(√3x+2) => x=√3/4 or/& x=-2/√3 ans.

Answer 2

$4\sqrt3\ x^2+5x-2\sqrt3=0\\\\a=4\sqrt3;\ b=5;\ c=-2\sqrt3\\\\\Delta=b^2-4ac\to\Delta=5^2-4(4\sqrt3)(-2\sqrt3)=25+96=121\\\\x=\frac{-b\pm\sqrt\Delta}{2a}\\\\x=\frac{-5\pm\sqrt{121}}{2(4\sqrt3)}=\frac{-5\pm11}{8\sqrt3}\Rightarrow x=\frac{-16}{8\sqrt3}\ or\ x=\frac{6}{8\sqrt3}\\\\x=\frac{-2}{\sqrt3}\cdot\frac{\sqrt3}{\sqrt3}=-\frac{2\sqrt3}{3}\ or\ x=\frac{3}{4\sqrt3}\cdot\frac{\sqrt3}{\sqrt3}=\frac{3\sqrt3}{4(3)}=\frac{\sqrt3}{4}\\\\Answer:\boxed{x=-\frac{2\sqrt3}{3}\ or\ x=\frac{\sqrt3}{4}}$