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Rationalizing Denominators Generator: Free Practice Clearing Radicals

Practice rationalizing a denominator, drilling monomial radicals, conjugate binomials, and higher-index roots, with instant feedback across three difficulty levels.

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Rationalizing Denominators Generator

The Rationalizing Denominators lesson covers every version of the rationalizing multiplier in detail; this generator drills the result those multipliers always produce — a whole-number denominator, checkable directly.

How the difficulty levels work

DifficultyWhat it drillsExample
EasyA monomial square-root denominator\(\dfrac{5}{\sqrt{7}}\), new denominator \(7\)
MediumA conjugate binomial denominator\(\dfrac{4}{3+\sqrt{5}}\), new denominator \(4\)
HardA bigger conjugate, or a cube-root denominator\(\dfrac{2}{\sqrt[3]{9}}\), new denominator \(27\)

Easy rationalizes a single square-root term in the denominator, where the new denominator is always exactly the original radicand.

Medium rationalizes a conjugate binomial denominator like \(a+\sqrt{b}\), where the new denominator always works out to \(a^{2}-b\).

Hard alternates between a bigger conjugate binomial and a cube-root denominator, where completing the cube always produces a denominator equal to a perfect cube.

Using the generator

Pick a difficulty and a question count, and a fresh set appears instantly. Every answer here is a whole number — enter it exactly as computed.

  • Check Answers scores the set and flags exactly which problems need another look
  • Show Answers reveals every solution, useful for reviewing a paper attempt
  • Print Worksheet outputs a clean page for offline practice or classroom handouts

The rules this generator drills

RuleStatement
Monomial denominator\(\dfrac{1}{\sqrt{a}} \cdot \dfrac{\sqrt{a}}{\sqrt{a}} = \dfrac{\sqrt{a}}{a}\)
Conjugate denominator\((a+\sqrt{b})(a-\sqrt{b}) = a^{2}-b\)
Higher-index denominatorMultiply by enough extra factors to reach a full \(n\)th power under the root

See the Rationalizing Denominators lesson for the full derivation of each multiplier and worked examples with the numerator included.

Common mistakes to watch for

Squaring a binomial denominator instead of using its conjugate. \((a+\sqrt{b})^{2}\) still contains a radical; only multiplying by the conjugate clears it.

Flipping the wrong sign when writing a conjugate. The conjugate of \(3-\sqrt{5}\) is \(3+\sqrt{5}\) — only the sign between the two terms changes.

Using an identical copy of the radical for a higher-index root. For \(\sqrt[3]{x}\), multiplying by another \(\sqrt[3]{x}\) only reaches \(x^{2}\) under the root, not a perfect cube.

Forgetting the multiplier has to equal 1. Whatever is multiplied onto the denominator has to be multiplied onto the numerator too — this generator only checks the denominator, but a complete answer always keeps both.

Where to go next

The exact same conjugate trick reappears in the Complex Number Arithmetic Generator for dividing by a complex number. For the full technique and every worked example, see the Rationalizing Denominators lesson.

Frequently Asked Questions

Why does every problem ask for the new denominator instead of the full rationalized fraction?+

A fully rationalized fraction like (2-√3)/1 is an expression with a radical still in the numerator, which can't be automatically checked as a single number. Rationalizing always turns the denominator into a whole number, though, so asking for that one piece is a completely reliable check.

How do I find the new denominator for a monomial radical, like 5/√7?+

Multiplying by √7/√7 always turns the denominator into exactly the radicand — 7 in this example — since √a times √a is a. No other calculation is needed for a simple square-root denominator.

How do I find the new denominator for a conjugate binomial, like 1/(3+√5)?+

Multiplying by the conjugate always produces a difference of squares in the denominator: (a+√b)(a-√b) = a² − b. For 3+√5 that's 3² − 5 = 4, without needing to expand the numerator at all.

Why does a cube-root denominator sometimes need more than one extra factor?+

A cube root needs its radicand raised to a full multiple of 3 to clear completely, so if the radicand is already r² under the root, only one more factor of r is needed; if it's just r, two more factors are needed. Either way, the resulting denominator is always r cubed.

Do I need Simplifying Radicals before this generator?+

It helps, since a denominator is usually simplified first in a full problem, but this generator's denominators are already in a form ready to rationalize directly, so it can be practiced on its own.

Is there a limit on how many problems I can generate?+

No. Sets are generated on demand and are unlimited and free, with no account required.

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