6.5 Rational Expressions, Rational Equations & Radical Equations
Key Takeaways
- A rational expression is a ratio of polynomials; it is undefined wherever the denominator equals zero, and those excluded values must be stated before simplifying.
- Simplify a rational expression only by cancelling common *factors* after factoring completely; never cancel individual terms across a sum.
- Adding or subtracting rational expressions requires a least common denominator, while multiplying requires none and dividing means multiplying by the reciprocal.
- Solving a rational equation by clearing denominators can introduce extraneous solutions, so every candidate answer must be checked against the original excluded values.
- Squaring both sides of a radical equation is not a reversible step, so radical equations always require substitution back into the original equation.
Rational Expressions, Rational Equations & Radical Equations
Quick Summary: A rational expression is a fraction whose numerator and denominator are polynomials. The blueprint requires arithmetic on them alongside polynomials, plus solving simple rational and radical equations in one variable. Two ideas carry the whole topic: you may only cancel factors, never terms; and both equation types can produce extraneous solutions, so checking is not optional.
Rational Expressions and Excluded Values
A rational expression has the form $\dfrac{P}{Q}$ where $P$ and $Q$ are polynomials and $Q \neq 0$. Because division by zero is undefined, any value of the variable making the denominator zero is an excluded value (also called a restriction).
Find excluded values by setting the original denominator equal to zero and solving:
| Expression | Set denominator $= 0$ | Excluded values |
|---|---|---|
| $\dfrac{5}{x - 3}$ | $x - 3 = 0$ | $x \neq 3$ |
| $\dfrac{x + 1}{x^2 - 9}$ | $x^2 - 9 = (x-3)(x+3) = 0$ | $x \neq 3,\ x \neq -3$ |
| $\dfrac{2x}{x^2 + 4x}$ | $x(x + 4) = 0$ | $x \neq 0,\ x \neq -4$ |
Determine restrictions before you simplify. In the third row, simplifying gives $\dfrac{2x}{x(x+4)} = \dfrac{2}{x+4}$, which looks defined at $x = 0$ — but the original expression was not, so $x \neq 0$ still holds.
Simplifying: Cancel Factors, Never Terms
Factor the numerator and denominator completely, then divide out common factors.
The single most common error in this topic: cancelling across addition. In $\dfrac{x + 5}{5}$ you may not cancel the fives to get $x$. The 5 in the numerator is a term, not a factor. Cancellation is only legal when the shared piece multiplies everything above and everything below.
Arithmetic With Rational Expressions
The rules mirror ordinary fraction arithmetic exactly.
| Operation | Rule | Needs a common denominator? |
|---|---|---|
| Multiply | $\dfrac{a}{b} \cdot \dfrac{c}{d} = \dfrac{ac}{bd}$ — factor first, cancel, then multiply | No |
| Divide | $\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \cdot \dfrac{d}{c}$ — multiply by the reciprocal | No |
| Add / Subtract | $\dfrac{a}{c} \pm \dfrac{b}{c} = \dfrac{a \pm b}{c}$ — rewrite over the LCD first | Yes |
Worked Example 1: Multiplication
Cancel $(x - 3)$ and $(x + 2)$ diagonally before multiplying — expanding first creates a needless mess.
Worked Example 2: Subtraction With an LCD
The LCD is $(x-2)(x+1)$. Rewrite each fraction, then subtract numerators:
Distribute the subtraction across the entire second numerator. Writing $3x + 3 - 5x - 10$ is the classic sign slip.
Solving Rational Equations
A rational equation sets rational expressions equal. Solve by multiplying every term by the LCD to clear denominators, then solve the resulting polynomial equation — and check.
The Four-Step Method
- State the excluded values from the original denominators.
- Multiply every term by the LCD.
- Solve the resulting linear or quadratic equation.
- Discard any solution that equals an excluded value — it is extraneous.
Worked Example 3: A Solution That Must Be Rejected
Step 1: $x \neq 4$.
Step 2: Multiply through by $(x - 4)$: $x = 4 + 3(x - 4)$.
Step 3: $x = 4 + 3x - 12$, so $x = 3x - 8$, giving $-2x = -8$ and $x = 4$.
Step 4: But $x = 4$ is excluded. The equation has no solution. A candidate who skipped step 4 would confidently choose a wrong answer that the test writers have supplied on purpose.
Proportions Are Rational Equations
When each side is a single fraction, cross-multiplication is a shortcut for the LCD step: $\dfrac{a}{b} = \dfrac{c}{d}$ becomes $ad = bc$. The extraneous-solution check still applies.
Solving Radical Equations
A radical equation has the variable under a root. Isolate the radical, then raise both sides to the matching power.
Why Checking Is Mandatory
Squaring is not reversible: $-3 \neq 3$, but $(-3)^2 = 3^2$. Squaring can therefore create solutions the original equation never had.
Worked Example 4: One Valid, One Extraneous
Square both sides: $x + 6 = x^2$. Rearrange: $x^2 - x - 6 = 0$, which factors as $(x - 3)(x + 2) = 0$, giving $x = 3$ and $x = -2$.
Check both in the original equation:
- $x = 3$: $\sqrt{3 + 6} = \sqrt{9} = 3$. True — valid.
- $x = -2$: $\sqrt{-2 + 6} = \sqrt{4} = 2$, but the right side is $-2$. False — extraneous.
The solution is $x = 3$ only. The principal square root is never negative, so any equation of the form $\sqrt{\ \cdot\ } = (\text{negative})$ has no solution at all.
Worked Example 5: Isolate First
Subtract 5 before squaring: $\sqrt{2x - 1} = 3$. Now square: $2x - 1 = 9$, so $2x = 10$ and $x = 5$. Check: $\sqrt{9} + 5 = 8$. Valid.
Squaring before isolating would produce $2x - 1 + 10\sqrt{2x-1} + 25 = 64$ — still solvable, but far harder.
For which value(s) of x is the rational expression (x + 5) / (x^2 - 2x - 15) undefined?
Which expression is equivalent to (2 / (x - 3)) + (5 / x), written over a single denominator?
A student solves the rational equation x / (x - 5) = 5 / (x - 5) + 2 and obtains x = 5. What is the correct conclusion?
What is the complete solution set of the radical equation sqrt(3x + 4) = x - 2?