7.5 Radical and Rational Equations and Functions

One Content Area, Two Function Families

Radical and rational equations: Creating radical and rational equations and functions in one variable, determining domain and range for radical and rational functions, graphing radical and rational functions, and simplifying radical and rational expressions and equations.

Worth 1 to 3 of the 20 AAF questions. The two families are grouped together because they share a defining feature: each has values of $x$ that are simply not allowed, and each can generate solutions that fail when checked.

Part One: Radical Equations

The Solving Procedure

1. Isolate the radical on one side.
2. Square both sides (or raise to the index power).
3. Solve the resulting equation.
4. Check every solution in the original equation. Not optional.

Solve $\sqrt{2x + 3} = 5$. Square: $2x + 3 = 25 \Rightarrow 2x = 22 \Rightarrow x = 11$ Check: $\sqrt{2(11)+3} = \sqrt{25} = 5$ ✓

Extraneous Solutions

Squaring is not a reversible operation — it destroys sign information. Both $3$ and $-3$ square to $9$, so squaring can introduce values that solve the squared equation but not the original.

Solve $\sqrt{x + 6} = x$.

Square: $x + 6 = x^2 \Rightarrow x^2 - x - 6 = 0 \Rightarrow (x-3)(x+2) = 0$

Candidates: $x = 3$ and $x = -2$.

Check $x = 3$: $\sqrt{9} = 3$ ✓ Check $x = -2$: $\sqrt{4} = 2$, but the right side is $-2$. 2 ≠ −2 ✗

So $x = -2$ is extraneous and the only solution is $x = 3$.

The underlying reason: the principal square root is never negative, so $\sqrt{\ }= x$ requires $x \ge 0$ from the outset. Any negative candidate is automatically suspect.

Radical Function Domain and Range

For an even index (square root, fourth root), the radicand must be non-negative:

$f(x) = \sqrt{x - 4}$ requires $x - 4 \ge 0$, so the domain is $x \ge 4$.

The range is $y \ge 0$, since the principal root is never negative.

$f(x) = \sqrt{2x + 8}$: $2x + 8 \ge 0 \Rightarrow x \ge -4$.

For an odd index (cube root), there is no restriction — the domain and range are all real numbers, since negatives have real cube roots.

Graph Shape

$y = \sqrt{x}$ starts at the origin and rises rightward, flattening as $x$ grows. Transformations shift the starting point:

$y = \sqrt{x-4} + 2$ starts at $(4, 2)$.

Part Two: Rational Expressions and Equations

Simplifying

Factor numerator and denominator, then cancel common factors — never common terms.

$\dfrac{x^2 - 9}{x^2 + 7x + 12} = \dfrac{(x+3)(x-3)}{(x+3)(x+4)} = \dfrac{x-3}{x+4}$, with $x \neq -3, -4$

The restrictions come from the original denominator and survive simplification. Even though $(x+3)$ cancels, $x = -3$ was never in the domain.

Cancelling terms is invalid:

$\dfrac{x + 5}{5}$ does not reduce to $x$. The 5 in the numerator is a term, not a factor.

Domain of a Rational Function

Set the denominator equal to zero and exclude those values.

$f(x) = \dfrac{x+1}{x^2 - 5x + 6}$

Factor: $(x-2)(x-3) = 0 \Rightarrow x = 2, 3$. Domain: all reals except $2$ and $3$.

Solving Rational Equations

Multiply through by the least common denominator, solve, then discard any solution that makes an original denominator zero.

Solve $\dfrac{3}{x-2} = \dfrac{x}{x-2} - 1$.

Multiply by $(x-2)$:

$3 = x - (x-2) \Rightarrow 3 = x - x + 2 \Rightarrow 3 = 2$

A false statement, so there is no solution.

Solve $\dfrac{x^2}{x-4} = \dfrac{16}{x-4}$.

Multiply by $(x-4)$: $x^2 = 16 \Rightarrow x = \pm 4$.

But $x = 4$ makes the denominator zero, so it is extraneous. Only $x = -4$.

This is the signature rational-equation trap, and $x = 4$ will always be offered as a choice.

Holes Versus Vertical Asymptotes

Both arise from denominator zeros, and the distinction is whether the factor cancels:

$f(x) = \dfrac{(x-2)(x+5)}{(x-2)(x-3)}$

• $(x-2)$ cancels → hole at $x = 2$
• $(x-3)$ does not → vertical asymptote at $x = 3$

Both values are excluded from the domain; they simply look different on the graph.

Horizontal Asymptotes

Compare the degrees of numerator ($n$) and denominator ($d$):

$f(x) = \dfrac{3x^2 + 1}{5x^2 - x}$ → degrees equal → $y = \tfrac{3}{5}$

The Unifying Discipline

Both families demand the same habit, and it is the one students skip under time pressure — except that these tests are untimed, so there is no excuse:

Identify the restricted values before solving, and check every candidate solution against the original equation.

For radicals the restriction comes from the radicand; for rationals it comes from the denominator. In both cases the test constructs items in which the discarded value is the most attractive answer choice.