Start out by squaring both sides to clear the root, then use good algebra moves to solve for x:

$$ \begin{align} 3^2 &= \left( \sqrt{x - 1} \right)^2 \\[5pt] 9 &= x - 1 \\[5pt] x &= 10 \end{align}$$

Start out by squaring both sides to clear the root, then use good algebra moves to solve for x:

$$ \begin{align} \left( \sqrt{x + 5} \right)^2 &= 0^2\\[5pt] x + 5 &= 0 \\[5pt] x &= -5 \end{align}$$

Start out by squaring both sides to clear the roots, then use good algebra moves to solve for x:

$$ \begin{align} \left( \sqrt{2x - 6} \right)^2 &= \left( \sqrt{3x - 14} \right)^2\\[5pt] 2x - 6 &= 3x - 14 \\[5pt] -x &= -8 \\[5pt] x &= 8 \end{align}$$

Here's a graph of $\sqrt{2x-6}$ and $\sqrt{3x - 14}$. They intersect at $x = 8$.

Start out by squaring both sides to clear the root, then use good algebra moves to solve for x:

$$ \begin{align} \left( \sqrt{8x} \right)^2 &= x^2 \\[5pt] 8x &= x^2 \\[5pt] x &= 0, \, 8 \\[5pt] \end{align}$$

Notice that in the second-to-last step, we divided by x, thus "dropping" a root, $x = 0.$ We always have to be wary of that.

Start out by squaring both sides to clear the root, then use good algebra moves to solve for x:

$$ \begin{align} \left( \sqrt{3 - 2x} \right)^2 &= \left( \sqrt{1 - 3x} \right)^2\\[5pt] 3 - 2x &= 1 - 3x \\[5pt] x &= -2 \end{align}$$

This equation contains both roots and numbers without any exponent (recall that a root is just a rational exponent). A good way to tackle this problem is to arrange to:

$$\sqrt{x - 6} = \sqrt{x + 9} - 3$$

Now square both sides:

$$ \begin{align} \left( \sqrt{x - 6} \right)^2 &= \left( \sqrt{x + 9} - 3 \right)^2 \\[5pt] x - 6 &= \left( \sqrt{x + 9} \right)^2 \\[5pt] &\phantom{000} - 2(3)\sqrt{x + 9} + 3^2 \tag{1} \\[5pt] x - 6 &= x + 9 - 6 \sqrt{x + 9} + 9 \\[5pt] 6 \sqrt{x + 9} &= 24 \tag{2} \\[5pt] \sqrt{x + 9} &= 4 \\[5pt] \left( \sqrt{x + 9} \right)^2 &= 4^2 \tag{3}\\[5pt] x + 9 &= 16 \\[5pt] x &= 7 \end{align}$$

In step (1) we multiplied binomials. Remember that (a+b)² = a² + 2ab + b². In step (2) we just gathered like terms and put the root on the left side. In step (3) we need to square both sides again. That's common in these kinds of problems. The rest is just a few simple algebra moves. Here is a graph of $y = \sqrt{x - 6}$ (black) and $y = \sqrt{x+9}+3$ (red). They intersect at $x = 7$.

First move the constant (4) to the right and cube both sides. From then it's just some algebra moves:

$$ \begin{align} \sqrt[3]{x - 6} &= -2 \\[5pt] \left( \sqrt[3]{x - 6} \right)^3 &= (-2)^3 \\[5pt] x - 5 &= -8 \\[5pt] x &= -2 \end{align}$$

Here is a graph of $y = \sqrt[3]{x - 6}$ (black) and $y = -2$ (red). They intersect at $x = -2$.

With a problem like this – a mix of roots and powers, it's often better to express the root and the power together as a rational exponent, like this:

$$(x - 3)^{\frac{2}{3}} = 4$$

Now raise both sides of the equation to the reciprocal power, $\frac{3}{2}.$

$$ \begin{align} \left((x - 3)^{\frac{2}{3}}\right)^{\frac{3}{2}} &= 4^{\frac{3}{2}} \\[5pt] x - 3 &= \left( 4^{\frac{1}{2}} \right)^3 \\[5pt] x - 3 &= 2^3 \\[5pt] x &= 8 + 3 \\[5pt] x &= 11 \end{align}$$

Raise both sides to the fourth power, then solve for $x$:

$$ \begin{align} \left( \sqrt[4]{2x+2} \right)^4 &= 3^4 \\[5pt] 2x + 2 &= 3^4 \\[5pt] 2x + 2 &= 81 \\[5pt] 2x &= 79 \\[5pt] x &= \frac{79}{2} \end{align}$$

Move the 5 to the left, then square both sides and solve for $x$:

$$ \begin{align} (x - 5)^2 &= \left[ (3x - 11)^{\frac{1}{2}} \right]^2 \\[5pt] x^2 - 10 x + 25 &= 3x - 11 \\[5pt] x^2 -13 x + 36 &= 0 \\[5pt] (x-9)(x-4) &= 0 \\[5pt] x &= 9, \, 4 \end{align}$$

If we plug the $x = 4$ solution back into the original equation we get $4-5 = \sqrt{3x - 11}$, but that would imply a negative square root, so we have to toss that solution. The only solution is $x = 9$. Here's a graph of the solution

Square both sides, then solve for $x$:

$$ \require{cancel} \begin{align} \left(\sqrt{2x + 102} \right)^2 &= x + 11 \\[5pt] 2x + 102 &= (x + 11)^2 \\[5pt] 2x + 102 &= x^2 + 22x + 121 \\[5pt] x^2 + 20x + 19 &= 0 \\[5pt] (x + 19)(x + 1) &= 0 \\[5pt] x &= \cancel{-19}, \, -1 \end{align}$$

We have to strike the $x = -19$ solution because the original problem would then read,

$$ \begin{align} \sqrt{2(-19) + 102} &= -19 + 11 \\[5pt] \sqrt{64} &= -8, \end{align}$$

which can't be true, so $x = -19$ is an extraneous solution that we have to toss out.

Square both sides, then solve for $x$:

$$ \begin{align} -x + 33 &= x^2 - 6x + 9 \\[5pt] x^2 - 5x -24 &= 0 \\[5pt] (x - 8)(x + 3) &= 0 \\[5pt] x &= \cancel{8}, \, -3 \end{align}$$

The solution $x = 8$ must be put aside because it would cause our root in the original problem to equal a negative number:

$$\sqrt{-8 + 33} = -5$$

Square both sides, then solve for $x$:

$$ \begin{align} -4x + 12 &= 4x^2 - 24x + 36 \\[5pt] 4x^2 - 20x + 24 &= 0 \\[5pt] x^2 - 5x + 6 &= 0 \\[5pt] (x - 6)(x + 1) &= 0 \\[5pt] x &= \cancel{6}, \, -1 \end{align}$$

$x = 6$ is an extraneous solution.

Square both sides, then solve for $x$:

$$ \begin{align} \left( \sqrt{3-x} \right)^2 = \left( \sqrt{x+7} + 2 \right)^2 \\[5pt] 3 - x &= x + 7 + 4 \sqrt{x+7} + 4 \\[5pt] 3 - x - x - 7 - 4 &= 4 \sqrt{x+7} \\[5pt] \frac{-2-8}{4} &= \sqrt{x+7} \\[5pt] -\frac{x}{2} - 2 &= \sqrt{x+7} \end{align}$$

Now square both sides again:

$$ \begin{align} \left( -\frac{x}{2} - 2 \right)^2 = \left( \sqrt{x+7} \right)^2 \\[5pt] \frac{x^2}{4} + 2x + 4 &= x + 7 \\[5pt] \frac{x^2}{4} + x - 3 &= 0 \\[5pt] x^2 + 4x - 12 &= 0 \\[5pt] (x+6)(x-2) &= 0 \\[5pt] x &= -6, \, 2 \end{align}$$

Now we need to check these possible solutions:

$$\sqrt{3+6} = \sqrt{1} + 3 \; \color{green}{\checkmark}$$

$$\sqrt{3-2} \ne \sqrt{9} + 2 \; \color{red}{X}$$

The only solution is $x = -6$

$$ \begin{align} 4x &= x^2 - 16x + 64 \\[5pt] x^2 - 20x + 64 &= 0 \\[5pt] (x-16)(x-4) &= 0 \\[5pt] x &= 4, \, \cancel{16} \end{align}$$

$x = 16$ is an extraneous solution.

In this case, let's raise each side of the equation to the 4^th power:

$$ \begin{align} (x-2)^{\frac{4}{2}} &= (28-2x)^{\frac{4}{4}} \\[5pt] (x-2)^2 &= 28 - 2x \\[5pt] x^2 - 4x + 4 + 2x -28 &= 0 \\[5pt] x^2 -2x -24 &= 0 \\[5pt] (x-6)(x+4) &= 0 \\[5pt] x = 6, \, -4 \end{align}$$

Now to check these solutions in the orignal equation:

$$ \begin{align} \sqrt{6-2} &= (28 - 12)^{\frac{1}{4}} \\[5pt] \sqrt{4} &= \sqrt[4]{16} \\[5pt] 2 &= 2 \; \color{green}{\checkmark} \end{align}$$

Now -4:

$$ \begin{align} \sqrt{-4-2} &= (28 - (-8))^{\frac{1}{4}} \\[5pt] \sqrt{-6} &= \sqrt[4]{36} \end{align}$$

The solution $x = -4$ does not check out. We can't make an imaginary root equal to a real root.

Let's move the -2 to the right, then cube both sides of the equation:

$$ \begin{align} \sqrt[3]{x^2+4} &= 2\\[5pt] x^2+4 &= 2^3 \\[5pt] x^2 + 4 - 8 &= 0 \\[5pt] x^2 &= 4 \\[5pt] x &= \pm 2 \end{align}$$

We should check these solutions. First $x = 2$:

$$ \begin{align} \sqrt[3]{2^2 + 4} - 2 &= 0 \\[5pt] \sqrt[3]{8} - 2 &= 0 \\[5pt] 2 - 2 &= 0 \; \color{green}{\checkmark} \end{align}$$

And $x = -2$:

$$\sqrt[3]{(-2)^2 + 4} - 2 = 0$$

That checks out in the same way; both roots are good.

First cube both sides:

$$ \begin{align} x^2 + \cancel{x} - 12 &= \cancel{x} + 52 \\[5pt] x^2 - 64 &= 0 \\[5pt] x^2 &= 64 \\[5pt] x^2 &= \pm 8 \end{align}$$

We should check these solutions. First $x = 8$:

$$ \begin{align} \sqrt[3]{8^2 + 8 - 12} &= \sqrt[3]{8 + 52} \\[5pt] \sqrt[3]{64 + 8 - 12} &= \sqrt[3]{60} \\[5pt] \sqrt[3]{60} &= \sqrt[3]{60} \; \color{green}{\checkmark} \end{align}$$

And $x = -8$:

$$ \begin{align} \sqrt[3]{(-8)^2 + 8 - 12} &= \sqrt[3]{-8 + 52} \\[5pt] \sqrt[3]{60} &= \sqrt[3]{46} \; \color{red}{X} \end{align}$$

So only $x = 8$ is a solution.

Move the 6 to the right, divide by 2 and cube both sides:

$$ \begin{align} \sqrt[3]{3x - 8} &= \frac{2-6}{2} \\[5pt] 3x - 8 &= (-2)^3 \\[5pt] 3x - 8 &= -8 \\[5pt] x &= 0 \end{align}$$

We should check this solution:

$$ \begin{align} 2 \sqrt[3]{3(0)-8} + 6 &= 2 \\[5pt] 2(-2) + 6 &= 2 \\[5pt] 2 &= 2 \; \; \color{green}{\checkmark} \end{align}$$

So $x = 0$ is a good solution.

First square both sides:

$$ \begin{align} 8x + 9 &= x^2 \\[5pt] x^2 - 8x - 9 &= 0 \\[5pt] (x - 9)(x + 1) &= 0 \\[5pt] x &= 9, \; -1 \end{align}$$

We should check these solutions:

$$ \begin{align} \sqrt{8(9) + 9} &= 9 \\[5pt] \sqrt{81} &= 9 \; \color{green}{\checkmark} \\[5pt] \sqrt{8(-1) + 9} &= -1 \\[5pt] \sqrt{1} \ne -1 \; \; \color{red}{X} \end{align}$$

So only the $x = 9$ solution is good.