Solve simple rational and radical equations in one variable and explain how extraneous solutions can arise, checking every solution (TN A1.A.REI.A.2).

What this topic is asking

Standard A1.A.REI.A.2 covers simple rational equations (a variable in a denominator) and radical equations (a variable under a root), with a specific warning: these methods can create extraneous solutions, values that appear during solving but do not satisfy the original equation. The graded skill is solving correctly and checking every solution.

Rational equations: clear the denominator

To solve an equation with the variable in a denominator, multiply every term by the least common denominator to clear fractions, then solve the resulting equation. Any value that would make an original denominator zero is excluded from the domain and must be rejected.

Radical equations: square, then check

To solve $\sqrt{\,\text{expression}\,} = \text{value}$, isolate the radical first, then square both sides. Squaring is the step that can manufacture a root, because both $a$ and $-a$ square to $a^2$, so the squared equation may have solutions the original rejects.

Why extraneous solutions arise

Extraneous solutions are not mistakes in your algebra; they are a side effect of two specific moves. Squaring is not reversible: $\sqrt{x} = -2$ has no solution (a principal square root is never negative), yet squaring gives $x = 4$, a value the original never allowed. Clearing a denominator can multiply away the very restriction that a value is illegal: if $x = 0$ was forbidden, multiplying by $x$ hides that. Because these steps change the solution set, the only safe response is to substitute every candidate back into the original equation and keep only those that work. This is exactly the reasoning A1.A.REI.A.2 wants you to be able to explain.

How TNReady examines this topic

• Numeric response. Solve a radical or rational equation and enter the valid solution.
• Multiple choice. Identify which candidate is extraneous, or how many valid solutions remain.
• Multiple select. Choose all valid solutions after rejecting extraneous ones.

A clarifying idea is that radical equations often become quadratics after squaring, so the factoring and quadratic-formula skills from the quadratic module are the tools you use to finish, then the domain check decides which roots survive.

Isolating the radical before squaring

The order of operations matters: you must get the radical alone on one side before squaring, or the square will not remove it. Consider $\sqrt{x + 4} + 3 = 7$. Squaring immediately would leave a messy expression, so first subtract $3$ to isolate the radical: $\sqrt{x + 4} = 4$. Now squaring gives $x + 4 = 16$, so $x = 12$, and the check $\sqrt{16} + 3 = 7$ confirms it. Squaring a sum like $(\sqrt{x + 4} + 3)$ would have produced a cross term and reintroduced a radical, defeating the purpose. Isolate first, square second, check third.

Try this

Q1. Solve $\sqrt{2x - 1} = 3$. [1 point]

• Cue. Square: $2x - 1 = 9 \Rightarrow x = 5$; check $\sqrt{9} = 3$, valid.

Q2. Solve $\dfrac{1}{x} + 2 = \dfrac{5}{x}$ and state any excluded value. [2 points]

• Cue. Multiply by $x$: $1 + 2x = 5 \Rightarrow x = 2$; excluded value is $x = 0$, and $2$ is valid.