Solve: \(4 = \sqrt{x}\)

Solve: \(\sqrt{3x - 4} = 2\)

Solve: \(\sqrt{2x + 5} + 11 = 6\)

Solve: \(\sqrt{x - 3} = 5\)

$$\begin{align}
& \sqrt{x - 3} = 5 \\
& x - 3 = 25 \\
& x = 28
\end{align}$$

Checking for extraneous solutions:

$$\begin{align}
& \sqrt{28 - 3} = 5 \\
\end{align}$$

Solve: \(\sqrt{x - 1} = 2\)

$$\begin{align}
& \sqrt{x - 1} = 2 \\
& x - 1 = 4 \\
& x = 5
\end{align}$$

Checking for extraneous solutions:

$$\begin{align}
& \sqrt{5 - 1} = 2
\end{align}$$

Solve: \(\sqrt{2x - 1} + 3 = 6\)

$$\begin{align}
& \sqrt{2x - 1} + 3 = 6 \\
& \sqrt{2x - 1} = 3 \\
& 2x - 1 = 9 \\
& 2x = 10 \\
& x = 5
\end{align}$$

Checking for extraneous solutions:

$$\begin{align}
& \sqrt{2 \cdot 5 - 1} + 3 = 6
\end{align}$$

Solve: \(\sqrt{6x + 2} = \sqrt{5x + 3}\)

$$\begin{align}
& \sqrt{6x + 2} = \sqrt{5x + 3} \\
& 6x + 2 = 5x + 3 \\
& x + 2 = 3 \\
& x = 1
\end{align}$$

Checking for extraneous solutions:

$$\begin{align}
& \sqrt{6 \cdot 1 + 2} = \sqrt{8} \\
& \sqrt{5 \cdot 1 + 3} = \sqrt{8}
\end{align}$$

Solve: \(\sqrt{x - 7} = 7 - \sqrt{x}\)

$$\begin{align}
& \sqrt{x - 7} = 7 - \sqrt{x} \\
& x - 7 = (7 - \sqrt{x})^2 \\
& x - 7 = 49 - 14\sqrt{x} + x \\
& 14\sqrt{x} = 49 + x - x + 7 \\
& 14\sqrt{x} = 56 \\
& \sqrt{x} = 4 \\
& x = 16
\end{align}$$

Checking for extraneous solutions:

$$\begin{align}
& \sqrt{16 - 7} = 3 \\
& 7 - \sqrt{16} = 3
\end{align}$$

Solve: \(\sqrt{x - 3} + \sqrt{x} = 3\)

$$\begin{align}
& \sqrt{x - 3} + \sqrt{x} = 3 \\
& \sqrt{x - 3} = 3 - \sqrt{x} \\
& x - 3 = (3 - \sqrt{x})^2 \\
& x - 3 = 9 - 6\sqrt{x} + x \\
& 6\sqrt{x} = 9 + x - x + 3 \\
& 6\sqrt{x} = 12 \\
& \sqrt{x} = 2 \\
& x = 4
\end{align}$$

Check for extraneous solutions:

$$\begin{align}
& \sqrt{4 - 3} + \sqrt{4} = 3
\end{align}$$

Solve: \(\sqrt{x - 7} = \sqrt{2}\left(\sqrt{x} - 2\right)\)

$$\begin{align}
& \sqrt{x - 7} = \sqrt{2}\left(\sqrt{x} - 2\right) \\
& x - 7 = 2\left(\sqrt{x} - 2\right)^2 \\
& x - 7 = 2\left(x - 4\sqrt{x} + 4\right) \\
& x - 7 = 2x - 8\sqrt{x} + 8 \\
& 8\sqrt{x} = 2x + 8 - x + 7 \\
& 8\sqrt{x} = x + 15 \\
& 64x = (x + 15)^2 \\
& 64x = x^2 + 30x + 225 \\
& 0 = x^2 - 34x + 225 \\
& (x - 25)(x - 9) = 0 \\
& x = 25\quad\text{or}\quad x = 9
\end{align}$$

Checking for extraneous solutions:

$$\begin{align}
& \sqrt{25 - 7} = \sqrt{18} = 3\sqrt{2} \\
& \sqrt{2}\left(\sqrt{25} - 2\right) = 3\sqrt{2} \\[1em]
& \sqrt{9 - 7} = \sqrt{2} \\
& \sqrt{2}\left(\sqrt{9} - 2\right) = \sqrt{2}
\end{align}$$

Solve: \(\sqrt{2x - 3} + \sqrt{x + 2} = 3\)

Solve: \(\sqrt{x + 1} = \dfrac{3}{\sqrt{x - 1}} - \sqrt{x - 1}\)

Solve: \(\sqrt{x^2 + 3x + 6} - 4 = 0\)

Solve: \(\sqrt{x^2 + 8x + 5} - 5 = 0\)

Solve: \(3 + \sqrt{5x + 6} = 12\)

Solve: \(\sqrt{3x - 7} + \sqrt{2x - 1} = 0\)

Solve: \(6 + 3w = \sqrt{2w + 12} + 2w\)

Solve: \(\sqrt{x + 7} - 5 = x\)

Solve: \(\sqrt{3t + 1} = \sqrt{t + 15}\)

Solve: \(\sqrt{x^2 - 4} = x - 2\)

Solve: \(\sqrt{2x + 1} + 5 = 0\)

No solution.

Solve: \(\sqrt{5x - 1} + 3 = x\)

\(x = 10\)

Solve: \(\sqrt{4x - 3} + \sqrt{3x + 7} = \sqrt{15x + 4}\)

Solve: \(\sqrt{3x + 1} = 4\)

Solve: \(\sqrt{7 - x} + 3 = 5\)

Solve: \(\sqrt[3]{x + 15} = 3\)

Solve: \(2\sqrt{x} = x\)

Solve: \(x^\frac{1}{4} + 4 = 7\)

Solve: \(\sqrt{3x + 4} = \sqrt{4x + 3}\)

Solve: \(\sqrt{5 + x} - \sqrt{4x + 9} = -2\)

Solve: \(4\sqrt[3]{x - 7} + 1 = 9\)

Solve: \(\sqrt{x + 1} = x - 1\)

Solve: \(\sqrt{x + 6} + 2 = x + 6\)

Solve: \(\sqrt{4x + 1} - 5 = 0\)

Solve: \(\sqrt{5z - 6} + 7 = 10\)

Solve: \(\sqrt[3]{2x} + 4 = 6\)

Solve: \(\sqrt{3x + 4} - 4 = 0\)

Solve: \(2 + \sqrt{x + 5} = 4\)

Solve: \(\sqrt{4x + 2} = \sqrt{3x + 4}\)