Answer:
A) don't see an x B) x=23 C) x=2; D) x=-1
Step-by-step explanation:
B) Cube both sides...[tex](\sqrt[3]{4+x})^3=3^3[/tex]---> 4+x=27
C) add [tex]\sqrt{12}[/tex] to both sides... [tex]\sqrt{3x^2}=\sqrt{12}[/tex]... square both sides...[tex](\sqrt{3x^2})^2=(\sqrt{12})^2[/tex]--> 3x^2=12... divide both sides by 3--> x^2=4---> x=2
D= [tex]\sqrt{18}-(x*\sqrt{2})=\sqrt{32}[/tex]... divide each side [tex]\sqrt{2}[/tex] ---> [tex](\sqrt{18}/\sqrt{2})- [(x*\sqrt{2})/\sqrt{2}]= \sqrt{32}/\sqrt{2}[/tex]---> [tex]\sqrt{9}-x= \sqrt{16}[/tex]... and because 9 and 16 are perfect squares our equation now reads---> 3-x=4---> -x=1---> x=-1
