b) x ² + 2(1+ √3)x + √3 = 0
$Δ'=(1+\sqrt{3})^2-\sqrt{3}=4+\sqrt{3}>0$
⇒$x_1= -(1+\sqrt{3})-\sqrt{4+\sqrt{3}}$
$x_2= -(1+\sqrt{3})+\sqrt{4+\sqrt{3}}$
c) 3x ² - (3+ √11)x + √11 = 0
⇔$(3x^2-3x)-(\sqrt{11}x-\sqrt{11}))=0$
⇔$3x(x-1)-\sqrt{11}(x-1)=0$
⇔$(x-1)(3x-\sqrt{11})=0$
⇔\(\left[ \begin{array}{l}x=1\\x=\frac{\sqrt{11}}{3}\end{array} \right.\)