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 \( \sqrt{x+1} = 3 \).

\( \begin{aligned} \require{color}
\sqrt{x+1}^2 &= 3^2 &\color{red} \text{square both sides} \\
x+1 &= 9 \\
x &= 8 \\
\text{LHS } &= \sqrt{8+1} &\color{red} \text{check} \\
&= \sqrt{9} \\
&= 3 \\
\text{LHS } &= \text{ RHS} \\
\therefore x &= 3 \\
\end{aligned} \\ \)

Question 2

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

\( \begin{aligned} \require{color}
\sqrt{x-1} &= 5 -3 \\
\sqrt{x-1} &= 2 \\
\sqrt{x-1}^2 &= 2^2 &\color{red} \text{square both sides} \\
x-1 &= 4 \\
x &= 5 \\
\text{LHS } &= \sqrt{5-1} + 3 &\color{red} \text{check} \\
&= \sqrt{4} + 3 \\
&= 2 + 3 \\
&= 5 \\
\text{LHS } &= \text{ RHS} \\
\therefore x &= 5 \\
\end{aligned} \\ \)

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.

\( \begin{aligned} \require{color}
\sqrt{x+2} &= 1-4 \\
\sqrt{x+2} &= -3 \\
\sqrt{x+2}^2 &= (-3)^2 &\color{red} \text{square both sides} \\
x+2 &= 9 \\
x &= 7\\
\text{LHS } &= \sqrt{7+2} + 4 &\color{red} \text{check} \\
&= \sqrt{9} + 4 \\
&= 3 + 4 \\
&= 7 \\
&\ne 1\\
\text{LHS } &\ne \text{ RHS} \\
\therefore &\text{ no solution} \\
\end{aligned} \\ \)

Question 4

Solve \( \sqrt{9x} – 5 = \sqrt{4x} \).

\( \begin{aligned} \require{color}
\sqrt{9}\sqrt{x} – 5 &= \sqrt{4}\sqrt{x} \\
3\sqrt{x} – 5 &= 2\sqrt{x} \\
\sqrt{x} &= 5 \\
\sqrt{x}^2 &= 5^2 &\color{red} \text{square both sides} \\
x &= 25 \\
\text{LHS } &= \sqrt{9 \times 25} – 5 &\color{red} \text{check} \\
&= \sqrt{9} \times \sqrt{25} – 5 \\
&= 3 \times 5 – 5 \\
&= 15 – 5 \\
&= 10 \\
\text{RHS } &= \sqrt{4 \times 25} \\
&= \sqrt{4} \times \sqrt{25} \\
&= 2 \times 5 \\
&= 10 \\
\text{LHS } &= \text{ RHS} \\
\therefore x &= 25 \\
\end{aligned} \\ \)

Question 5

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

\( \begin{aligned} \require{color}
\sqrt{2x+1}^2 &= (x-1)^2 &\color{red} \text{square both sides} \\
2x+1 &= x^2 – 2x + 1 \\
x^2 – 4x &= 0 \\
x(x-4) &= 0 \\
x &= 0 \text{ or } 4 \\
\text{check for } x&= 0 \\
\text{LHS } &= \sqrt{2 \times 0 + 1} \\
&= \sqrt{1} \\
&= 1 \\
\text{RHS } &= 0-1 \\
&= -1 \\
\text{LHS } &\ne \text{ RHS for } x = 0 \\
\text{check for } x&= 4 \\
\text{LHS } &= \sqrt{2 \times 4 + 1} \\
&= \sqrt{9} \\
&= 3 \\
\text{RHS } &= 4-1 \\
&= 3 \\
\text{LHS } &= \text{ RHS for } x=4 \\
\therefore x &= 4 \\
\end{aligned} \\ \)

Question 6

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

\( \begin{aligned} \require{color} \displaystyle
\sqrt{x+10}^2 &= \sqrt{2}^2(x-5)^2 &\color{red} \text{square both sides} \\
x + 10 &= 2(x^2 -10x + 25) \\
x + 10 &= 2x^2 -20x + 50) \\
2x^2 -21x + 40 &= 0 \\
(2x-5)(x-8) &= 0 \\
x &= \frac{5}{2} \text{ or } 8 \\
\text{check for } x&= \frac{5}{2} \\
\text{LHS } &= \sqrt{\frac{5}{2}+10} \\
&= \sqrt{\frac{25}{2}} \\
&= \frac{5}{\sqrt{2}} \\
&= \frac{5\sqrt{2}}{2} \\
\text{RHS } &= \sqrt{2}(\frac{5}{2}-5) \\
&= \sqrt{2} \times \frac{-5}{2} \\
&= \frac{-5\sqrt{2}}{2} \\
\text{LHS } &\ne \text{ RHS for } x = \frac{5}{2} \\
\text{check for } x&= 8 \\
\text{LHS } &= \sqrt{8+10} \\
&= \sqrt{18} \\
&= 3\sqrt{2} \\
\text{RHS } &= \sqrt{2}(8-5) \\
&= \sqrt{2} \times 3 \\
&= 3\sqrt{2} \\
\text{LHS } &= \text{ RHS for } x = 8 \\
\therefore x &= 8 \\
\end{aligned} \\ \)

Question 7

Solve \( 3x – \sqrt{x} = 2 \).

\( \begin{aligned} \require{color} \displaystyle
3x – 2 &= \sqrt{x} \\
(3x – 2)^2 &= \sqrt{x}^2 &\color{red} \text{square both sides} \\
9x^2 -12x +4 &= x \\
9x^2 -13x +4 &= 0 \\
(9x-4)(x-1) &= 0 \\
x &= \frac{4}{9} \text{ or } 1 \\
\text{check for } x &= \frac{4}{9} \\
\text{LHS } &= 3 \times \frac{4}{9} – \sqrt{\frac{4}{9}} \\
&= \frac{4}{3} – \frac{2}{3} \\
&= \frac{2}{3} \\
&\ne 2 \\
\text{LHS } &\ne \text{ RHS for } x = \frac{4}{9} \\
\text{check for } x &= 1 \\
\text{LHS } &= 3 \times 1 – \sqrt{1} \\
&= 3 – 1 \\
&= 2 \\
\text{LHS } &= \text{ RHS for } x = 1 \\
\therefore x &= 1 \\
\end{aligned} \\ \)

Question 8

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

\( \begin{aligned} \require{color}
\Big(\sqrt{x+1} + \sqrt{2x}\Big)^2 &= 7^2 &\color{red} \text{square both sides} \\
x+1 + 2\sqrt{x+1}\sqrt{2x} + 2x &= 49 \\
3x+1 + 2\sqrt{2x(x+1)} &= 49 \\
2\sqrt{2x(x+1)} &= 48 – 3x \\
2^2\sqrt{2x(x+1)}^2 &= (48 – 3x)^2 &\color{red} \text{square both sides} \\
4 \times 2x(x+1) &= 2304 – 288x + 9x^2 \\
x^2 – 296x + 2304 &= 0 \\
(x-8)(x-288) &= 0 \\
x &= 8 \text{ or } 288 \\
\text{check for } x &= 8 \\
\text{LHS } &= \sqrt{8+1} + \sqrt{2 \times 8} \\
&= \sqrt{9} + \sqrt{16} \\
&= 3 + 4 \\
&= 7 \\
\text{LHS } &= \text{ RHS for } x=8 \\
\text{check for } x &= 288 \\
\text{LHS } &= \sqrt{288+1} + \sqrt{2 \times 288} \\
&= \sqrt{289} + \sqrt{576} \\
&= 17 + 24 \\
&= 41 \\
&\ne 7 \\
\text{LHS } &\ne \text{ RHS for } x=288 \\
\therefore x &= 8 \\
\end{aligned} \\ \)

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 🙂

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