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\(\begin{align}
 3x + 2 &\ge 0 \\
 3x &\ge - 2 \\
 x &\ge - \frac{2}{3}, x \in \rm R \\
 \end{align}\)

\(\begin{align}
 \sqrt {3x + 2} + 6 &= 4 \\
 \sqrt {3x + 2} &= -2 \\
 \left( {\sqrt {3x + 2} } \right)^2 &= \left( { - 2} \right)^2  \\
 3x + 2 &= 4 \\
 3x &= 2 \\
 x &= \frac{2}{3} \\
 \end{align}\)

The solution is within the variable’s restrictions.

Verify:

Left Side	Right Side
\(\begin{array}{r}  \sqrt {3x + 2} + 6 \\  \sqrt {3\left( {\frac{2}{3}} \right) + 2} + 6 \\  \sqrt {2 + 2} + 6 \\  \sqrt 4 + 6 \\  2 + 6 \\  8 \\  \end{array}\)	\(4\)
LS \(\ne\) RS \(\hspace{30pt}\)

Because the left side does not equal the right side, \(x = \frac{2}{3}\) does not satisfy the equation and is considered an extraneous root. There is no solution to this equation.

\(\begin{align}
 r - 5 &\ge 0 \\
 r &\ge 5, r \in \rm R \\
 \end{align}\)

\(\begin{align}
 \sqrt[4]{{r - 5}} &= 3 \\
 \left( {\sqrt[4]{{r - 5}}} \right)^4 &= 3^4  \\
 r - 5 &= 81 \\
 r &= 86 \\
 \end{align}\)

The solution is within the variable’s restrictions.

Verify:

Left Side	Right Side
\(\begin{array}{r}  \sqrt[4]{{r - 5}} \\  \sqrt[4]{{86 - 5}} \\  \sqrt[4]{{81}} \\  3 \\  \end{array}\)	\(3\)
LS = RS

The solution is \(r = 86\).