I’m getting tangled up with rationalising denominators. I know the goal is “no square roots in the denominator,” but I keep second-guessing what I’m supposed to multiply by and when I’ve actually finished.
Like, for something simple-looking like 7/√5, I think I multiply by √5… but then for 4/(√2√3), do I multiply by √6, or treat them separately? And for 5/(2+√3), I’ve seen people use the conjugate (2−√3), but I don’t fully get why that’s the right move. Same with 1/(√3−√5) – is it always the conjugate? What if the denominator is 3√2 – is that still considered “not rationalised,” even though there’s a 3 hitching a ride?
Tiny tangent: does this change for cube roots? Does the conjugate idea still apply, or is that a different trick altogether?
I keep ending up with another surd popping back into the denominator or making the expression messier. Could someone explain a simple rule-of-thumb for choosing what to multiply by in each case and how to tell when it’s properly rationalised?
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3 Responses
I think of it as: multiply by whatever makes the denominator a perfect square/power-so 7/√5 × √5, 4/(√2√3) × √6 (same as √2 then √3), and for sums/differences like 5/(2+√3) or 1/(√3−√5) use the conjugate to get a difference of squares; 1/(3√2) still isn’t “done,” so I’d clear the √2, and for cube roots the analogue is a^3−b^3, i.e., multiply (a±b) by a^2∓ab+b^2.
I’m pretty sure that’s the rule-of-thumb, though I still double-check the signs on that last identity and occasionally wonder if leaving a 3√2 is “good enough.”
Pattern: pair each irrational factor with what makes a perfect power-single √a: multiply by √a; product √2√3 (= √6): multiply by √6; sums/differences use the conjugate (a±√b → a∓√b, or √a±√b → √a∓√b) so cross-terms cancel; 3√2 is still irrational, so multiply by √2; for cube roots use the “cubic partner” x^2 ∓ xy + y^2 so (x±y)(x^2 ∓ xy + y^2) = x^3 ± y^3-done when the denominator is rational (no radicals). Example: 5/(2+√3) × (2−√3)/(2−√3) = 5(2−√3)/(4−3) = 5(2−√3).
Rule of thumb: multiply by the smallest factor that turns the entire denominator into a rational number. For a single square root like 7/√5, multiply top and bottom by √5 to get 7√5/5. If the denominator is a product of square roots, combine first: 4/(√2√3) = 4/√6, then multiply by √6 to get 2√6/3; doing √2 and √3 separately amounts to the same thing. If there’s a constant times a root, like 1/(3√2), it’s still not rationalized; multiply by √2 to get √2/6. For sums or differences with square roots, use the conjugate because (a + b√m)(a − b√m) = a^2 − b^2m is rational: 5/(2 + √3) × (2 − √3)/(2 − √3) gives denominator 4 − 3 = 1, and 1/(√3 − √5) × (√3 + √5)/(√3 + √5) gives denominator 3 − 5 = −2. You’re “done” when the denominator has no radicals and you’ve simplified any common factors. For cube roots, the idea shifts: you still aim for a perfect cube. A lone ∛m needs ∛(m^2) (e.g., 1/∛5 → ∛25/5). For binomials with a cube root, use the factor that completes a sum/difference of cubes: if the denominator is a + b∛m, multiply by a^2 − ab∛m + b^2∛(m^2) so the denominator becomes a^3 + b^3m (rational); for example, 1/(2 + ∛3) times (4 − 2∛3 + ∛9) gives denominator 8 + 3 = 11. When I learned this, I made a quick checklist: is it a single root (make a perfect square/cube), a product (combine first), or a sum/difference (use the appropriate conjugate). That small routine stopped the second-guessing and kept the algebra tidy.