A numerical radical in a denominator is removed by multiplying the fraction by a carefully chosen form of 1. For one term the multiplier is the radical itself; for two terms it is the conjugate, and the difference of squares does the rest. The value never changes — that is what makes the move legal.
Core ideas
The technique is one idea used three ways, and the idea is that multiplying by 1 is always allowed.
√7/√7 is 1, so (3/√7)·(√7/√7) is the same number as 3/√7. This is the same licence used to build a common denominator in “Adding and Subtracting Rational Expressions”; nothing is being added to or removed from the expression.
For √7 you need one more √7 to make 7. For ∛2 you need ∛4, because ∛2 · ∛4 = ∛8 = 2. The question is always: what does this radicand need in order to become an exact nth power?
Simplify and combine first. If two unlike terms remain, multiplying √5 − √2 by √5 + √2 gives 5 − 2 = 3 by the difference of squares, and the denominator is rational in one step.
The material
Three shapes of denominator: one square root, one higher-index root, and a two-term sum.
Why bother? Two reasons, and neither is that the unrationalized form is wrong. 3/√7 is a perfectly good number. But 3√7/7 is easier to estimate by hand, and — the reason that survives — a single agreed form lets two answers be compared. Three students with 3/√7, 3√7/7 and (3/7)√7 have written the same number three ways.
To rationalize a denominator is to rewrite the fraction as an equal fraction whose denominator contains no radical.
Multiply top and bottom by that radical. 3/√7 = (3·√7)/(√7·√7) = 3√7/7, since √7 · √7 = 7. The radical did not disappear — it moved to the numerator, where the convention allows it.
Here the reflex fails. For 1/∛2, multiplying by ∛2/∛2 gives ∛2/∛4, which is no better. You need the denominator to become a perfect CUBE, so multiply by ∛4/∛4: 1/∛2 = ∛4/∛8 = ∛4/2. Ask what the radicand is missing, not what it already is.
For a denominator a + √b, multiply top and bottom by a − √b; the new denominator is a² − b, which contains no radical.
For √a + √b the conjugate is √a − √b and the new denominator is a − b. Both are the difference of two squares, and squaring is what defeats a square root.
6/(√5 − √2). Multiply above and below by √5 + √2. The denominator becomes (√5)² − (√2)² = 5 − 2 = 3, and the fraction becomes 6(√5 + √2)/3 = 2(√5 + √2). Both sides evaluate to 7.3006….
For the numerical denominators treated here, simplify and combine before choosing the multiplier. Then the displayed form of 1 is a nonzero number and the rewrite is reversible. For example, √2 + √2 is first 2√2; its apparent conjugate √2 − √2 is zero and is not a form of 1. Symbolic denominators whose multipliers may vanish need separate domain analysis and are outside this lesson.
One warning about signs. The conjugate of √5 − √2 is √5 + √2, and the conjugate of 3 − √2 is 3 + √2. Only the sign BETWEEN the terms flips; flipping the sign of the whole expression instead gives −(√5 − √2), whose product with the original is −(5 − 2√10 + 2), still full of radicals.
Interactive
One term by the root, two by the conjugate
Method
Identify the shape of the denominator first; the shape names the multiplier.
Extract perfect powers and reduce the fraction first. Rationalizing a denominator that was about to simplify anyway creates work and larger numbers.
One radical term, or two terms with a radical? A single ⁿ√ takes the completing multiplier; a two-term denominator takes the conjugate.
For ⁿ√(cᵏ) multiply by ⁿ√(cⁿ⁻ᵏ) so the exponents sum to n. For a ± √b multiply by a ∓ √b, changing only the sign between the terms.
The denominator should now be rational; expand the numerator fully and cancel any common factor. 6(√5 + √2)/3 is not finished until it is 2(√5 + √2).
A square root, a cube root, and a conjugate pair.
Step (b) is the one that separates the technique from the reflex. Multiplying by ∛2/∛2 is a legal move that accomplishes nothing; the denominator must be completed to ∛(2³), so the missing factor is ∛4.
In (c) the final cancellation matters. 6(√5 + √2)/3 and 2(√5 + √2) are the same number, but only the second is in simplified form, and the 3 in the denominator came from the conjugate product rather than from anything in the original fraction.
For a faded choice of multiplier, rationalize both 2/∛9 and 4/(3 + √5). The supplied questions are: what factor completes 9 to a cube, and what sign makes the second denominator a difference of squares? Work from those decisions to 2∛3/3 and 3 − √5. If the first still has a cube root below, you multiplied by what was already there instead of the missing factor; if the second denominator is 14 + 6√5, you squared a binomial instead of using its conjugate.
Given any of the three shapes you should be able to name the form of 1 before doing any arithmetic, and say what the denominator will become. You should also be able to say why this step, unlike the one in “Solving Radical Equations”, can never affect a solution set.
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