Radical and Rational Equations

Radical and rational equations are algebraic equations that require specific techniques to isolate the variable. Radical equations involve variables inside a root (like a square root), while rational equations involve variables in the denominator of a fraction.

When solving both types of equations, the methods we use can sometimes create extraneous solutions—answers that emerge from the algebra but do not actually make the original equation true. Because of this, checking your final answers is a mandatory step.

Solving Radical Equations

To solve an equation with a radical, the goal is to eliminate the root. You do this by isolating the radical on one side of the equation and then raising both sides to a power equal to the index of the root (e.g., squaring both sides for a square root).

Example: Solve $\sqrt{3x + 1} = x - 1$

1. Square both sides to eliminate the square root: $(\sqrt{3x + 1})^2 = (x - 1)^2$ $3x + 1 = x^2 - 2x + 1$

2. Rearrange into a standard polynomial equation (set to zero): $0 = x^2 - 5x$

3. Solve for $x$: $x(x - 5) = 0$ This gives potential solutions of $x = 0$ and $x = 5$.

4. Check for extraneous solutions by plugging them into the original equation:

   • For $x = 0$: $\sqrt{3(0) + 1} = 0 - 1 \implies \sqrt{1} = -1$. Since the principal square root of $1$ is $1$, and $1 \neq -1$, $x = 0$ is an extraneous solution.
   • For $x = 5$: $\sqrt{3(5) + 1} = 5 - 1 \implies \sqrt{16} = 4$. This is true.

The only valid solution is $x = 5$.

Solving Rational Equations

To solve a rational equation, the most efficient method is to clear all fractions by multiplying every term on both sides by the Lowest Common Denominator (LCD).

Example: Solve $\frac{2}{x - 1} + \frac{3}{x + 2} = 1$

1. Identify the LCD and domain restrictions. The denominators are $(x - 1)$ and $(x + 2)$. The LCD is $(x - 1)(x + 2)$. Note: $x$ cannot be $1$ or $-2$, as these would cause division by zero.

2. Multiply the entire equation by the LCD: $(x - 1)(x + 2) \left[ \frac{2}{x - 1} + \frac{3}{x + 2} \right] = 1 \cdot (x - 1)(x + 2)$

3. Simplify and expand: $2(x + 2) + 3(x - 1) = (x - 1)(x + 2)$ $2x + 4 + 3x - 3 = x^2 + x - 2$ $5x + 1 = x^2 + x - 2$

4. Set the equation to zero and solve: $0 = x^2 - 4x - 3$ Since this doesn't factor neatly, use the quadratic formula: $x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-3)}}{2(1)}$ $x = \frac{4 \pm \sqrt{16 + 12}}{2} = \frac{4 \pm \sqrt{28}}{2}$ $x = \frac{4 \pm 2\sqrt{7}}{2} = 2 \pm \sqrt{7}$

5. Check against restrictions: Neither $2 + \sqrt{7}$ nor $2 - \sqrt{7}$ equals $1$ or $-2$. Therefore, both are valid solutions.

Summary of Key Rules

• Isolate & Power: For radicals, isolate the root first, then apply the exponent.
• Clear Fractions: For rational equations, multiply everything by the LCD.
• Always Check: Squaring both sides or multiplying by a variable expression can introduce fake (extraneous) answers. Always verify solutions in the original equation.