5.
Given the equation to solve for x:
[tex]3(x+1)^{\frac{2}{3}}=12[/tex]The steps for the solution are as follows:
[tex]\begin{gathered} 3(x+1)^{\frac{2}{3}}=12 \\ \frac{3(x+1)^{\frac{2}{3}}}{3}=\frac{12}{3} \\ (x+1)^{\frac{2}{3}}=4 \\ \lbrack(x+1)^{\frac{2}{3}}\rbrack^{\frac{1}{2}}=(4)^{\frac{1}{2}} \\ \lbrack(x+1)^{\frac{1}{3}}\rbrack=\pm2 \\ \lbrack(x+1)^{\frac{1}{3}}\rbrack^3=(\pm2)^3 \\ x+1=\pm8 \end{gathered}[/tex]From the above equation, we have x + 1 = 8 and x + 1 = -8. These imply x = 7 and x = -9.
Check for extraneous solutions:
If x = 7, then the left-hand side of the equation is:
[tex]3(x+1)^{\frac{2}{3}}=3(7+1)^{\frac{2}{3}}=3(4)=12[/tex]Thus, the equation holds true at x = 7.
If x = -9, then the right-hand side of the equation is:
[tex]3(x+1)^{\frac{2}{3}}=3(-9+1)^{\frac{2}{3}}=3(4)=12[/tex]Thus, the equation holds true at x = -9.
There is no extraneous solution. The solutions of the given equation are x = 7 and x = -9.
6.
Given an equation to solve for x:
[tex]\sqrt[]{3x+2}-2\sqrt[]{x}=0[/tex]The steps of the solution are as follows:
[tex]\begin{gathered} \sqrt[]{3x+2}-2\sqrt[]{x}=0 \\ \sqrt[]{3x+2}=2\sqrt[]{x} \\ (\sqrt[]{3x+2})^2=(2\sqrt[]{x})^2 \\ 3x+2=4x \\ 2=4x-3x \\ 2=x \end{gathered}[/tex]Thus, the solution of the equation is x = 2.
