# 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 of both sides is to be squared but not squaring individual terms.
This is what I say,
\large \begin{align} (1+2)^2 &= 3^2 \\ 1^2 + 2^2 &\ne 3^2 \end{align}

### Checkpoint 2 for Solving radical equations

Make sure to check the solutions in the original equation.
\large \begin{aligned} \require{AMSsymbols} \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{AMSsymbols} \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{AMSsymbols} \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 explains why the solutions must be checked for solving radical equations.
\begin{aligned} \require{AMSsymbols} \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{AMSsymbols} \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{AMSsymbols} \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}\left(\frac{5}{2}-5\right) \\ &= \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{AMSsymbols} \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{AMSsymbols} \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 $\sqrt{x+5} + \sqrt{x-2} = \sqrt{5x-6}$.

Please let us know if you need further help solving this equation. Have fun ðŸ™‚