Solving Radical Equations

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 🙂