If you remember one rule from the radicals chapter, it is probably \sqrt{a}\cdot\sqrt{b} = \sqrt{ab}. It lets you pull apart \sqrt{50} into \sqrt{25}\cdot\sqrt{2} = 5\sqrt{2}, and it feels so natural that a tempting question follows: does the same trick work with a plus sign? Is \sqrt{a} + \sqrt{b} equal to \sqrt{a+b}?
It is not. And the asymmetry between multiplication and addition — why the square root splits over one but not the other — is a place where a lot of students secretly lose marks. So let us settle it once, with a numerical check that cannot be argued with, and then an algebraic reason that explains the check.
The thirty-second numerical check
Before any theory, try the rule on a pair of numbers you already know.
Why these numbers: 9 and 16 are perfect squares, so both sides are integers — no approximation, no rounding, no room to hide a small error.
Left side: \sqrt{9 + 16} = \sqrt{25} = 5.
Right side: \sqrt{9} + \sqrt{16} = 3 + 4 = 7.
5 \neq 7. Different numbers. The "rule" fails on the first example you try, and it fails by a lot — not by a rounding error. So the identity \sqrt{a+b} = \sqrt{a} + \sqrt{b} is flatly false.
The comparable check for multiplication:
Left side: \sqrt{144} = 12. Right side: 3 \cdot 4 = 12. Same.
So the multiplication rule survives the check and the addition rule dies. That is the fact. Now for the reason.
Why multiplication lets the root split
The rule \sqrt{a}\cdot\sqrt{b} = \sqrt{ab} is not arbitrary — it follows from the definition of \sqrt{\,\,}.
A square root is a number whose square is the radicand. So (\sqrt{a}\cdot\sqrt{b})^2 had better equal ab if we want \sqrt{a}\cdot\sqrt{b} to qualify as \sqrt{ab}. Let us check:
Why the rearrangement: multiplication is commutative, so we can shuffle \sqrt{a}\sqrt{b}\sqrt{a}\sqrt{b} into (\sqrt{a})^2(\sqrt{b})^2. Then each (\sqrt{\,})^2 collapses back to the radicand.
So \sqrt{a}\cdot\sqrt{b} is indeed a number whose square is ab, and since both \sqrt{a} and \sqrt{b} are non-negative (by the principal-root convention), their product is non-negative too, which matches the principal-root convention for \sqrt{ab}. So \sqrt{a}\cdot\sqrt{b} = \sqrt{ab} exactly.
The rule works because squaring a product is the product of squares: (xy)^2 = x^2 y^2. Multiplication and squaring cooperate.
Why addition refuses
Now try the same argument with a plus sign. Suppose — for contradiction — that \sqrt{a} + \sqrt{b} = \sqrt{a+b}. Square both sides:
Expand the left side using (x + y)^2 = x^2 + 2xy + y^2:
Why the extra term: (x+y)^2 has a cross term 2xy that (xy)^2 does not. Squaring a sum produces extra middle terms; squaring a product does not. This 2\sqrt{ab} is the culprit.
For the equality to hold, we would need a + b + 2\sqrt{ab} = a + b, i.e., 2\sqrt{ab} = 0. That forces ab = 0 — at least one of a or b must be zero. In every other case, the left side is strictly larger than the right.
So \sqrt{a} + \sqrt{b} \geq \sqrt{a+b}, with equality only when one of them is zero. In the 9, 16 example, the excess is exactly 2\sqrt{9 \cdot 16} = 2 \cdot 12 = 24; squaring the right side gives 7^2 = 49, squaring the left side gives 5^2 = 25, and the difference 49 - 25 = 24 matches exactly.
The addition rule fails because squaring a sum introduces a cross term, and the square root cannot "undo" that cross term. Multiplication has no cross term, so the root splits cleanly.
A second way to see it: the shape of the function
The square-root function y = \sqrt{x} is concave. Geometrically, its graph bows upward, meaning the chord between two points on the curve lies below the curve. Formally, for any positive a, b:
Multiplying both sides by 2 rearranges to \sqrt{2(a+b)} \geq \sqrt{a}+\sqrt{b}, which is stronger than what we need. Concavity is the reason sums under a square root compress rather than split — the function grows slowly, so "combining first, rooting after" gives less than "rooting first, combining after."
You cannot feel this from an algebraic identity alone, but it is consistent with the quick number check: \sqrt{9+16} compressed to 5, while \sqrt{9} + \sqrt{16} stayed spread out at 7. The extra space between 5 and 7 is the concavity in action.
The same pattern for every operation
The split-vs-no-split distinction is not special to square roots. Every function worth studying respects multiplication-ish structure better than addition-ish structure — or vice versa — and you can usually check which with the same squaring trick.
| Function | Splits over multiplication? | Splits over addition? |
|---|---|---|
| Square root, \sqrt{\,\,} | Yes (\sqrt{ab} = \sqrt{a}\sqrt{b}) | No |
| Squaring, x^2 | Yes ((ab)^2 = a^2 b^2) | No ((a+b)^2 \neq a^2 + b^2) |
| Logarithm, \log | Converts multiplication to addition | No (\log(a+b) \neq \log a + \log b) |
| Reciprocal, 1/x | Yes (\tfrac{1}{ab} = \tfrac{1}{a}\cdot\tfrac{1}{b}) | No (\tfrac{1}{a+b} \neq \tfrac{1}{a} + \tfrac{1}{b}) |
The pattern is consistent: the operations that interact smoothly with exponent laws (multiplication, division, powers) split, and the operation that introduces cross terms when squared (addition) does not. Anytime you find yourself wanting to split a function over a sum, check first with a perfect-square example — the first sum you try will probably betray the rule.
The correct moves near a sum under a root
You cannot split \sqrt{a+b}, but you can sometimes factor first. If a + b shares a common factor, pull it out:
In the second case, the sum evaluates first, and then the product rule pulls the perfect square out. The rule is: add under the root first, split out perfect-square factors second. Never split a sum directly.
For nested radicals with sums inside sums — expressions like \sqrt{3 + 2\sqrt{2}} — there are more advanced denesting techniques (see How to Handle Nested Radicals Like √(3 + 2√2)), but even there, the first step is never "split the sum."
The fastest way to remember
Every time the temptation to write \sqrt{a+b} = \sqrt{a}+\sqrt{b} comes back, run the 9, 16 check in your head: left gives 5, right gives 7, not equal. Ten seconds. That one example is a permanent immunisation.
And if you ever catch yourself doubting the multiplication rule — "wait, does \sqrt{ab} really split?" — run the 9 \cdot 16 check: left gives 12, right gives 12, equal. The two examples together are the cleanest way to keep the rules straight: multiplication splits, addition does not, and a single pair of perfect squares confirms both in under a minute.
Related: Roots and Radicals · How to Handle Nested Radicals Like √(3 + 2√2) · Rationalise the Denominator: Watch the Conjugate Flush the Irrational Upstairs · Exponents and Powers