Solving radical equations are required to isolate the radicals or surds to one side of the equations. Then square both sides. Here we have two important checkpoints.

### Checkpoint 1 for Solving radical equations

Make sure the whole both sides are to be squared, but not squaring individual terms.
This is what I say,
$$(1+2)^2 = 3^2 \\ 1^2 + 2^2 \ne 3^2$$

### Checkpoint 2 for Solving radical equations

Make sure to check the solutions in the original equation.
\begin{aligned} \require{color} \sqrt{x} &= -2 \\ \sqrt{x}^2 &= (-2)^2 &\color{green} \text{squaring both sides} \\ x &= 4 \\ \text{LHS} &= \sqrt{4} &\color{green} \text{check} \\ &= 2 \\ &\ne -2 \\ \text{LHS } &\ne \text{ RHS} \\ \therefore &\text{ no solution} \\ \end{aligned} \\
OK, now let’s try to attempt the following questions.

### Question 1

Solve $$\sqrt{x+1} = 3$$.

### Question 2

Solve $$\sqrt{x-1} + 3 = 5$$.

### Question 3

Solve $$\sqrt{x+2} + 4 = 1$$.
Note that this question gives you the reason why the solutions have to be checked for solving radical equations.

### Question 4

Solve $$\sqrt{9x} – 5 = \sqrt{4x}$$.

### Question 5

Solve $$\sqrt{2x+1} = x-1$$.

### Question 6

Solve $$\sqrt{x+10} = \sqrt{2}(x-5)$$.

### Question 7

Solve $$3x – \sqrt{x} = 2$$.

### Question 8

Solve $$\sqrt{x+1} + \sqrt{2x} = 7$$.

Hope these example questions are useful for you to understand better about solving radical equations.

### Question for you

Solve $$\sqrt{x+5} + \sqrt{x-2} = \sqrt{5x-6}$$.

Feel free to let us know if you need further helps on solving this equation. Have fun 🙂