Question

√6x+√24=0

(2-√3)x-2√3=0

Answer 1

$\sqrt{6}x+ \sqrt{24}= 0 \\ \sqrt{6}x=- \sqrt{24} \\ x= -\frac{ \sqrt{24} }{ \sqrt{6} } =- \sqrt{ \frac{24}{6} }=- \sqrt{4}=-2 \\ (2- \sqrt{3})x-2 \sqrt{3}=0 \\ (2- \sqrt{3})x=2 \sqrt{3} \\ x= \frac{2 \sqrt{3}}{2- \sqrt{3}} = \frac{(2+ \sqrt{3})*2 \sqrt{3}}{(2+ \sqrt{3})(2- \sqrt{3})} = \frac{4 \sqrt{3}+6 }{4-3}= 4 \sqrt{3}+6$