SAT radical and rational equations: solve, then check for extraneous answers
Squaring both sides or clearing denominators can create answers that do not actually work. Learn the solve-then-check routine, see exactly where a false solution fails, and practice with rational exponents.
SPSAT Practice Editorial8 min read
The short version
What you'll take away
- Isolate the radical, square both sides, solve, and then check every candidate in the original equation.
- A candidate that fails the check is extraneous. It came from squaring, not from the original equation.
- In rational equations, any value that makes a denominator zero is excluded, even if the algebra produces it.
In this article
Why squaring can create false answers
Squaring both sides of an equation is a useful step, but it loses information about signs. Both 2 and −2 square to 4, so an equation like \(\displaystyle 2 = -2\), which is false, becomes \(\displaystyle 4 = 4\), which is true. Squaring can turn a false statement into a true one, so it can introduce solutions the original equation never had.
Those extra values are called extraneous solutions. The fix is simple: check every candidate in the original equation.
Solve, then check
Solve \(\displaystyle \sqrt{x + 7} = x + 1\). The radical is already alone, so square both sides: \(\displaystyle x + 7 = x^2 + 2x + 1\). Rearrange to \(\displaystyle x^2 + x - 6 = 0\), which factors as \(\displaystyle (x + 3)(x - 2) = 0\). The candidates are −3 and 2.
See the skill
Check each candidate in √(x + 7) = x + 1
| Candidate | Substitute into the original | Verdict |
|---|---|---|
| x = 2 | √(2 + 7) = 3 and 2 + 1 = 3 | ✓ Solution |
| x = −3 | √(−3 + 7) = 2 but −3 + 1 = −2 | ✗ Extraneous |
Rational equations: watch the denominators
To solve an equation with variables in denominators, multiply both sides by the common denominator. For \(\displaystyle \frac{6}{x} + 1 = 4\): subtract 1 to get \(\displaystyle \frac{6}{x} = 3\), so \(\displaystyle x = 2\). Check: \(\displaystyle \frac{6}{2} + 1 = 4\).
Now try \(\displaystyle \frac{x}{x - 3} = \frac{3}{x - 3} + 2\). Multiplying by \(\displaystyle x - 3\) gives \(\displaystyle x = 3 + 2(x - 3)\), so \(\displaystyle x = 2x - 3\) and \(\displaystyle x = 3\). But \(\displaystyle x = 3\) makes both denominators zero, so it is not allowed. The equation has no solution.
Rational exponents
A fractional exponent combines a root and a power: \(\displaystyle x^{m/n} = \sqrt[n]{x^m} = (\sqrt[n]{x})^m\). Taking the root first usually keeps numbers small. For example, \(\displaystyle 8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4\), and \(\displaystyle x^{1/2}\) is the same as \(\displaystyle \sqrt{x}\).
Your turn
Quick practice
What are all the solutions to √(2x + 3) = x?
Hint: Find both candidates, then substitute each one into the original equation.
Practice: try these before checking the answers
Check every candidate in the original equation.
- 1. Solve \(\displaystyle \sqrt{x - 1} = 3\).
- 2. Solve \(\displaystyle \sqrt{x + 5} = x - 1\).
- 3. Solve \(\displaystyle \frac{2}{x + 1} = \frac{1}{2}\).
- 4. Evaluate \(\displaystyle 27^{2/3}\).
- 5. Solve \(\displaystyle \frac{x}{x - 4} = \frac{4}{x - 4}\).
Sources & further reading
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