# 3.4 - Solving Rational and Radical Equations

## Radical Equations
- **Radical Equations**. (These equations involve $\sqrt{\cdots}$ or $\sqrt[n]{\cdots}$)
- Basic principle: If $\sqrt{X}=a$, then $X=a^2$. We can see this as 'squaring both sides': if

$$\sqrt{X}=a$$
then
$$(\sqrt{X})^2=a^2$$

$$X=a^2$$

---

### Example 1. Solve $\sqrt{3x+1}=4$

#### Solution:
Identify what's happening to $x$ according to PEMDAS, then do the opposite things in reverse order.

##### What's happening to $x$?
1. Multiplication by 3 to get $3x$.
2. Addition by 1 to get $3x+1$.
3. Square root to get $\sqrt{3x+1}$.

##### How do we undo this?
1. Square.
2. Subtract by 1.
3. Divide by 3.

| Step | Operation |
|------------------------------|----------------------------------|
| $\sqrt{3x + 1} = 4$ | Original Equation |
| $\left(\sqrt{3x + 1}\right)^2 = (4)^2$ | **Square both sides** |
| $3x+1 = 16$ | Simplify |
| $3x+1-1 = 16-1$ | **Subtract 1 from both sides** |
| $3x = 15$ | Simplify |
| $\dfrac{3x}{3} = \dfrac{15}{3}$ | **Divide both sides by 3** |
| $x = 5$ | Simplify |

**Final Answer:**

$$x=5$$

**Check $x=5$:**

\begin{align}
\sqrt{3x+1}&=4 \\
\sqrt{3(5)+1}&=4 \\
\sqrt{15+1}&=4 \\
\sqrt{16}&=4 \\
4&=4 \quad \checkmark \\
\end{align}

---

### Example 2. Solve $5+\sqrt{x+7}=10$

| Step | Operation |
|------------------------------|-------------------------------------|
| $5 + \sqrt{x + 7} = 10$ | Original Equation |
| $\sqrt{x + 7} = 10 - 5$ | Subtract 5 from both sides |
| $\sqrt{x + 7} = 5$ | Simplify |
| $\left( \sqrt{x + 7} \right)^2 = 5^2$ | Square both sides |
| $x + 7 = 25$ | Simplify |
| $x + 7 - 7 = 25 - 7$ | Subtract 7 from both sides |
| $x = 18$ | Simplify |

**Final Answer:**

$$x = 18$$

**Check $x=18$:**

\begin{align}
5+\sqrt{x+7}&=10 \\
5+\sqrt{18+7}&=10 \\
5+\sqrt{25}&=10 \\
5+5&=10 \\
10&=10 \quad \checkmark \\
\end{align}

---

- Exception. If $a<0$, you accomplish nothing by squaring both sides, because $\sqrt{X}$ cannot be negative. So if $\sqrt{X}=a<0$, then there is NO SOLUTION for $X$.

### Example 3. Solve $10+\sqrt{2x+5}=6$

#### Step 1: Isolate the square root term
Start by isolating the square root on one side of the equation:
$$
10 + \sqrt{2x + 5} = 6
$$
Subtract $10$ from both sides:
$$
\sqrt{2x + 5} = 6 - 10
$$
$$
\sqrt{2x + 5} = -4
$$

#### Step 2: Recognize the impossibility
The equation $\sqrt{2x + 5} = -4$ is impossible because the square root of any real number is always non-negative (positive or zero), and $-4$ is negative.

#### Conclusion:
Since the square root cannot be negative, the equation has **no real solution**.

##### Solving the normal way and then checking:

\begin{align}
10 + \sqrt{2x + 5} &= 6 \\
\sqrt{2x + 5} &= -4 \\
(\sqrt{2x+5})^2&=(-4)^2 \\
2x+5&=16 \\
2x&=11 \\
x&=11/2 \\
x&=5.5
\end{align}

**Check $x=5.5$:**

\begin{align}
10 + \sqrt{2(5.5) + 5} &= 6 \\
10 + \sqrt{11 + 5} &= 6 \\
10 + \sqrt{16} &= 6 \\
10 + 4 &= 6 \\
14&=6 \quad \text{ False.}\\
\end{align}

$x=5.5$ is not a solution. No solution.

## Rational equations. (These equations involve fractions, or `ratios'.)

- Basic principle: Multiply by the least common denominator of the fractions in the equation, which will cancel all denominators, and eliminate fractions.

---

### Example 4. $\displaystyle{\frac{2}{x}+\frac{1}{3}=\frac{1}{4}+\frac{3}{x}}$

The least common denominator is the product of the denominators in this case: $(x)(3)(4)=12x$. Multiply by this:

| Equation | Operation |
|---------------------------------------------|----------------------------------------------|
| $\frac{2}{x} + \frac{1}{3} = \frac{1}{4} + \frac{3}{x}$ | Original Equation |
| $12x\left(\frac{2}{x}\right) + 12x\left(\frac{1}{3}\right) = 12x\left(\frac{1}{4}\right) + 12x\left(\frac{3}{x}\right)$ | Multiply both sides by $12x$ |
| $12(2) + 4x(1) = 3x(1) + 12(3)$ | Cancel common factors |
| $24 + 4x = 3x + 36$ | Simplify |
| $24 + x = 36$ | Subtract $3x$ from both sides |
| $x = 36 - 24$ | Subtract $24$ from both sides |
| $x = 12$ | Simplify |

**Final Answer:**

$$x = 12$$

---

### Example 5. Solve $\frac{2}{3}x + \frac{5}{2} = \frac{1}{2}x + 3$

The least common denominator is $(3)(2)=6$:

| Equation | Operation |
|-------------------------------------------------------------|------------------------------------------|
| $\frac{2}{3}x + \frac{5}{2} = \frac{1}{2}x + 3$ | Original Equation |
| $6\cdot \frac{2}{3}x + 6\cdot \frac{5}{2} = 6\cdot \frac{1}{2}x + 6\cdot 3$ | Multiply both sides by the least common denominator $6$ |
| $4x + 15 = 3x + 18$ | Simplify |
| $4x - 3x = 18 - 15$ | Subtract $3x$ from both sides, subtract $15$ from both sides |
| $x = 3$ | Simplify |

**Final Answer:**

$$x = 3$$

---

### Example 6. Solve $\frac{3}{x-2} = \frac{4}{x+4}$

The least common denominator is $(x-2)(x+4)$, so

| Equation | Operation |
|-------------------------------------------------------------|------------------------------------------|
| $\frac{3}{x-2} = \frac{4}{x+4}$ | Original Equation |
| $(x-2)(x+4)\cdot \frac{3}{x-2} = (x-2)(x+4)\cdot \frac{4}{x+4}$ | Multiply both sides by the least common denominator $(x-2)(x+4)$ |
| $3(x+4) = 4(x-2)$ | Cancel common factors |
| $3x + 12 = 4x - 8$ | Expand both sides |
| $12 + 8 = 4x - 3x$ | Move like terms to one side |
| $x = 20$ | Simplify |

**Final Answer:**

$$x = 20$$

---

### Example 7. Solve $x - \frac{12}{x} = 1$

The least common denominator is $x$, so:

| Equation | Operation |
|-------------------------------------------------------------|------------------------------------------|
| $x - \frac{12}{x} = 1$ | Original Equation |
| $x\left(x - \frac{12}{x}\right) = x(1)$ | Multiply both sides by $x$ (LCD is $x$) |
| $x(x) - x\frac{12}{x} = x(1)$ | Distribute $x$ |
| $x^2 - 12 = x$ | Cancel common factors |
| $x^2 - x - 12 = 0$ | Subtract $x$ from both sides |
| $(x - 4)(x + 3) = 0$ | Factor the quadratic |
| $x = 4$ or $x = -3$ | Solve each factor |

**Final Answer:**

$$x = 4 \quad \text{or} \quad x = -3$$

---

### Example 8. Solve $\frac{4}{2x+1} = \frac{3}{x+3}$

The least common denominator is $(2x+1)(x+3)$.
| Equation | Operation |
|-------------------------------------------------------------|------------------------------------------|
| $\frac{4}{2x+1} = \frac{3}{x+3}$ | Original Equation |
| $(2x+1)(x+3)\cdot \frac{4}{2x+1} = (2x+1)(x+3)\cdot \frac{3}{x+3}$ | Multiply both sides by the least common denominator $(2x+1)(x+3)$ |
| $4(x+3) = 3(2x+1)$ | Cancel common factors |
| $4x + 12 = 6x + 3$ | Expand both sides |
| $12 - 3 = 6x - 4x$ | Move like terms to one side |
| $9 = 2x$ | Simplify |
| $x = \frac{9}{2}$ | Divide by 2 |

**Final Answer:**

$$x = \frac{9}{2}$$

---

### Example 9. Solve $\frac{1}{t} + 2 = \frac{3}{2t} - 3$
The least common denominator is $2t$.

| Equation | Operation |
|-------------------------------------------------------------|------------------------------------------|
| $\frac{1}{t} + 2 = \frac{3}{2t} - 3$ | Original Equation |
| $2t\left(\frac{1}{t} + 2\right) = 2t\left(\frac{3}{2t} - 3\right)$ | Multiply both sides by the least common denominator $2t$ |
|$(2t) \left(\dfrac{1}{t}\right)+(2t)(2)=(2t)\left(\dfrac{3}{2t}\right)+(2t)(-3)$ | Distribute|
|$\dfrac{2t}{1} \cdot \dfrac{1}{t}+2t \cdot 2=\dfrac{2t}{1} \cdot \dfrac{3}{2t}+2t(-3)$ | Put $2t$ over 1. |
| $2 + 4t = 3 - 6t$ | Simplify |
| $4t + 6t = 3 - 2$ | Move like terms to one side |
| $10t = 1$ | Simplify |
| $t = \frac{1}{10}$ | Divide both sides by 10 |

**Final Answer:**

$$t = \frac{1}{10}$$

---

### Example 10. Solve $\frac{1}{x+3} + \frac{2}{x-3} = \frac{4}{x^2-9}$

The least common denominator is $(x+3)(x-3)=x^2-9$.[^1]

[^1]: By FOILing:\begin{align}(x+3)(x-3)&=(x)(x)+(x)(-3)+(3)(x)+(3)(-3) \\
&=x^2-3x+3x-9 \\
&=x^2-9
\end{align}

| Equation | Operation |
|-------------------------------------------------------------|------------------------------------------|
| $\frac{1}{x+3} + \frac{2}{x-3} = \frac{4}{x^2-9}$ | Original Equation |
| $\frac{1}{x+3} + \frac{2}{x-3} = \frac{4}{(x+3)(x-3)}$ | Factor $x^2-9=(x+3)(x-3)$ |
| $(x+3)(x-3)\left(\frac{1}{x+3} + \frac{2}{x-3}\right) = (x+3)(x-3)\cdot \frac{4}{(x+3)(x-3)}$ | Multiply both sides by $x^2 - 9 = (x+3)(x-3)$ |
| $(x+3)(x-3)\cdot \dfrac{1}{x+3} + (x+3)(x-3)\cdot \dfrac{2}{x-3} = (x+3)(x-3)\cdot \frac{4}{(x+3)(x-3)}$ | Simplify |
| $(x-3) + 2(x+3) = 4$ | Simplify |
| $x - 3 + 2x + 6 = 4$ | Expand |
| $3x + 3 = 4$ | Simplify |
| $3x = 1$ | Subtract 3 from both sides |
| $x = \frac{1}{3}$ | Divide by 3 |

**Final Answer:**

$$x = \frac{1}{3}$$

---

### Example 11. Solve $\frac{1}{x-2} - \frac{1}{x+2} = \frac{4}{x^2 - 4}$

The least common denominator is $(x+2)(x-2)=x^2-4$.[^2]

[^2]: By FOILing:\begin{align}(x+2)(x-2)&=(x)(x)+(x)(-2)+(2)(x)+(2)(-2) \\
&=x^2-2x+2x-4 \\
&=x^2-4
\end{align}

Sometimes all the $x$s disappear, as we saw with linear equations:

| Equation | Operation |
|-------------------------------------------------------------|------------------------------------------|
| $\frac{1}{x-2} - \frac{1}{x+2} = \frac{4}{x^2 - 4}$ | Original Equation |
| $(x^2-4)\left(\frac{1}{x-2} - \frac{1}{x+2}\right) = (x^2-4)\cdot \frac{4}{x^2 - 4}$ | Multiply both sides by $x^2 - 4 = (x-2)(x+2)$ |
| $(x+2) - (x-2) = 4$ | Simplify |
| $x + 2 - x + 2 = 4$ | Expand |
| $4 = 4$ | Simplify |

**Final Answer:**

Since we got a true statement, the solution is all real numbers.

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