1
Like radicals are like terms
3√2 + 5√2 = 8√2 for the same reason 3k + 5k = 8k. The radical is just a number you have not evaluated, and the distributive law does the work: 3√2 + 5√2 = (3 + 5)√2.
Lesson · Rational and Radical Expressions · Radicals
Adding, Subtracting and Multiplying Radicals
Addition of radicals is the collecting of like terms wearing a disguise: only radicals with the same index and the same radicand may be combined, and you cannot tell which those are until everything is simplified. Multiplication is friendlier, and the conjugate pair is the case worth memorising.
√8 + √18 = 2√2 + 3√2 = 5√2 simplify FIRST, then combine
√2 + √3 ≠ √5
(2 + √3)(2 − √3) = 4 − 3 = 1 conjugates: the root disappears
(√5 + √2)² = 7 + 2√10 the middle term does not
What combines and what does not
Core ideas
What this lesson establishes
Three rules govern the arithmetic, and two of them are rules about when you are NOT allowed to do the obvious thing.
2
Simplify before deciding what is alike
√8 and √18 look unrelated and are both multiples of √2. Any judgement about which radicals combine, made before simplifying, is a guess.
3
Multiplication has no such restriction
√2 · √3 = √6 needs nothing to match, only the product rule and non-negative radicands. Sums are the difficult direction, which is why the conjugate trick — turning a sum into a product — is worth so much in “Rationalizing Denominators”.
The material
Combining radicals, and the two ways it goes wrong
Addition first, because it is the one with a condition; then multiplication and the two special products.
Definition · Like radicals
Two radicals are like if they have the same index and the same radicand: 3√2 and −7√2 are like, √2 and √3 are not, and √2 and ∛2 are not.
Theorem · Adding like radicals
a·ⁿ√c + b·ⁿ√c = (a + b)·ⁿ√c.
This is the distributive law with ⁿ√c in the role of a common factor. There is no corresponding rule for unlike radicals, because there is no common factor to take out.
So √2 + √3 is already in simplest form. It cannot be written as a single radical, and in particular it is not √5: numerically √2 + √3 ≈ 3.1463 while √5 ≈ 2.2361. An expression being unfinished to the eye does not mean there is a step left.
Example · Alike after simplifying
√8 + √18 looks like two unlike radicals. Simplify: √8 = 2√2 and √18 = 3√2, so the sum is 5√2. The same expression, one line later, is a single term.
Multiplying
The product rule handles it: multiply the outside parts, multiply the radicands, then simplify what results. 3√2 · 4√6 = 12√12, and √12 is not simplified — 12 = 4 · 3, so 12√12 = 12 · 2√3 = 24√3. The last simplification is the step most often skipped.
With sums, distribute exactly as with polynomials. Nothing new happens; the terms are simply irrational.
Theorem · The conjugate product
(a + √b)(a − √b) = a² − b, and more generally (√a + √b)(√a − √b) = a − b for a, b ≥ 0.
It is the difference of two squares from Polynomials and Factoring, and squaring is exactly what removes a square root. The result contains no radical at all — which is the whole reason “Rationalizing Denominators” works.
Example · Two products, one with a middle term and one without
(2 + √3)(2 − √3) = 4 − 3 = 1: the cross terms −2√3 and +2√3 cancel. But (√5 + √2)² = 5 + 2√10 + 2 = 7 + 2√10: here the cross terms are equal and ADD. Squaring a binomial never loses its middle term.
That contrast is worth holding on to. (a + b)² = a² + b² is false for numbers and stays false for radicals, and “Solving Radical Equations” turns on it: squaring √x + 1 gives x + 2√x + 1, so squaring an equation with two terms on a side does not remove the root.
Interactive
Adding radicals
Like radicals only, once everything is simplified
Method
Working an expression with several radicals
Simplification comes first in every case. Until it is done you cannot see what you have.
1
Simplify every radical in the expression
Extract perfect powers, reduce indices, and clear fractions from under the signs. Only now is it visible which radicals are alike.
2
Group the like radicals and add their coefficients
Same index and same radicand. Treat each distinct radical the way you would treat a distinct variable when collecting like terms.
3
For products, distribute and use the product rule
Multiply coefficients together and radicands together. FOIL applies unchanged to two binomials; a conjugate pair is the case where the middle terms cancel.
4
Simplify the result, then stop
A product often produces a new radicand with a perfect power in it. Leave unlike radicals side by side — an answer such as 7 + 2√10 is finished.
Combine or leave
Score 0 / 0Choose the answer you can justify.
Four expressions worth the practice
Two sums and two products. Watch which step is doing the work.
(a) √8 + √18
= 2√2 + 3√2 both simplify to multiples of √2
= 5√2 check: 2.8284… + 4.2426… = 7.0710…
(b) √50 − √18
= 5√2 − 3√2 = 2√2 check: 7.0710… − 4.2426… = 2.8284…
(c) 3√2 · 4√6
= 12√12 multiply outsides, multiply radicands
= 12 · 2√3 = 24√3 12 = 4 · 3, so √12 = 2√3
check: 41.5692… both ways
(d) (√5 + √2)²
= (√5)² + 2·√5·√2 + (√2)² square of a binomial, middle term kept
= 5 + 2√10 + 2
= 7 + 2√10 check: 13.3245… both ways
In (c) the answer 12√12 is correct and unfinished. A radical answer is not simplified until nothing extractable remains under the sign, and √12 still hides a 4.
Compare (d) with (2 + √3)(2 − √3) = 1. Same shape of expression, opposite outcome: with a conjugate the middle terms cancel and the root vanishes, with a square they reinforce and the root survives. “Rationalizing Denominators” uses the first; “Solving Radical Equations” is made difficult by the second.
For a faded operation pass, first simplify √12 + 2√27 − √75; the supplied start is √12 = 2√3. Finish the other two extractions and combine to 3√3. Then expand (√6 + √2)(√6 − √2) from the supplied recognition that the factors are conjugates; the result is 4. If the first answer is √12, you combined radicands, and if the second retains a middle term, you gave the cross terms the same sign.
Common mistakes
1
Adding the radicands√50 − √18 is not √32. Radicals add like terms, not like numbers inside a box; simplify each one and combine the coefficients.
2
Judging likeness before simplifying√8 and √18 are alike and do not look it. Every decision about combining belongs after the simplification step, never before.
3
Losing the middle term when squaring(√5 + √2)² is 7 + 2√10, not 7. The cross terms cancel only for a conjugate PAIR, where the signs differ.
Check your understanding
✓
Finish when you can say, for any two radical terms, whether they combine and why.
Given an expression mixing sums and products you should be able to simplify every term, combine exactly the like ones, and leave the rest alone without discomfort. 7 + 2√10 is a finished answer, and treating it as unfinished is what produces an invented step.
Note: The conjugate product proved here is used immediately. “Rationalizing Denominators” needs a multiplier that clears a two-term radical denominator, and the difference of squares is the only thing that does it in one step.