You are halfway through a problem. A \sqrt{a + b} shows up. Your pen, without asking permission, writes:
It looks so clean. The square-root symbol sits in front of a sum; you "distribute" it over the two terms; you walk away feeling productive. And you are wrong. Not slightly wrong, not wrong in a corner case — wrong almost every single time this identity gets written. Examiners have been collecting this mistake for fifty years; it sits comfortably in the top five algebra errors on standardised tests, right next to (a+b)^2 = a^2 + b^2 (see the sibling article on algebra tiles).
This article is about why the move is illegal, what distribution rule you do have for square roots, and how to recognise the same mistake in all its disguises.
Test it with numbers
You do not need a theorem to catch this one. Pick a = 9 and b = 16 — two perfect squares so the arithmetic is clean.
The identity fails on its very first numerical test. In any proof, in any exam paper, that single counterexample is enough to mark the statement wrong. You do not need to explain why — you have produced two numbers for which the two sides disagree, and "for all a and b" is dead.
Any time you are tempted to write \sqrt{a + b} = \sqrt{a} + \sqrt{b}, run this check in your head. Pick easy numbers, compute both sides, see them disagree, and put the pen down.
Why it fails — the square root is nonlinear
A function f is called linear when it satisfies
for every a and every b. Linear functions are the only ones that "distribute over addition" — that is what linearity means. The classic examples are f(x) = 3x and f(x) = x/5: their graphs are straight lines through the origin, and you can pull apart a sum inside the function.
The square root f(x) = \sqrt{x} is not linear. Its graph is not a straight line — it is a curve that rises steeply near 0 and flattens out as x grows. Curving functions do not distribute over addition; the whole point of the curve is that the output at a + b is not the same as the outputs at a and b added together.
If the square-root graph were a straight line through the origin, the identity would hold. It is not, so it does not.
What DOES distribute over √
The square root is perfectly well-behaved over multiplication. For any a \geq 0 and b \geq 0,
You can check this with the same numbers. \sqrt{9 \cdot 16} = \sqrt{144} = 12, and \sqrt{9} \cdot \sqrt{16} = 3 \cdot 4 = 12. They agree. They will always agree, for every non-negative a and b — because squaring both sides gives a \cdot b = a \cdot b, which is obviously true.
So the rule is not "no distribution at all." The rule is: the square root distributes over multiplication, never over addition. Memorise the distinction. It is one of the sharpest lines in algebra.
Why multiplication works and addition doesn't
Watch what happens when you square each "candidate identity" and see which one is true.
The multiplication case. Is \sqrt{a} \cdot \sqrt{b} equal to \sqrt{a b}? Square both sides and see.
The left side squared is a b. The right side squared is also a b. They match, so the identity is true.
The addition case. Is \sqrt{a} + \sqrt{b} equal to \sqrt{a + b}? Square the left side.
This is a + b + 2\sqrt{ab}, not a + b. The two sides disagree by exactly 2\sqrt{ab} — a leftover term that refuses to go away.
That 2\sqrt{ab} is the missing piece. It is what makes \sqrt{a + b} genuinely different from \sqrt{a} + \sqrt{b}, and it is structurally identical to the 2ab that goes missing in (a+b)^2 = a^2 + b^2. The cross-term is always where the mistake lives.
Where the identity would be true
The only way to kill the 2\sqrt{ab} term is to make it zero. That means \sqrt{ab} = 0, which means a b = 0, which means a = 0 or b = 0.
So the identity \sqrt{a + b} = \sqrt{a} + \sqrt{b} is true only in the trivial case where one of the numbers is zero: \sqrt{0 + 5} = \sqrt{0} + \sqrt{5} is fine because \sqrt{0} = 0. For any two positive a and b, the identity fails. There is no positive pair where it holds. None.
Same family of distribution errors
The "distribute over addition" bug is not specific to square roots. It infects every nonlinear operation a student meets. Here is the full rogues' gallery.
- (a + b)^2 \neq a^2 + b^2 — the freshman's dream, missing 2ab (see sibling article).
- \dfrac{1}{a + b} \neq \dfrac{1}{a} + \dfrac{1}{b} — try a = b = 1: left is \tfrac{1}{2}, right is 2.
- \log(a + b) \neq \log a + \log b — but \log(a \cdot b) = \log a + \log b does hold. The log, like the square root, distributes over multiplication, not addition.
- \sin(a + b) \neq \sin a + \sin b — the real identity is \sin(a+b) = \sin a \cos b + \cos a \sin b, another cross-term story.
- e^{a + b} \neq e^a + e^b — but e^{a + b} = e^a \cdot e^b does hold, which is the exponential function's multiplication distribution law.
The pattern is unmissable. Nonlinear operations never distribute over addition. Only linear operations — scalar multiplication, the identity function, differentiation, integration — distribute over addition. Every time you see a function wrapped around a sum, stop and ask: is this function linear? If not, you cannot pull it apart.
What to do when you actually want to split √(a+b)
In general, you cannot. The expression \sqrt{a + b} is often stuck the way it is, and the best move is to leave it alone and work around it. But three techniques sometimes rescue you.
Factor first. If the expression inside the root has a common factor, pull it out and then take its square root.
You have not split the 1 + x; you have only extracted the 4. The \sqrt{1 + x} part stays as one piece.
Complete a square. If the sum inside is secretly a perfect square, rewrite it and take the root exactly.
Here the addition inside was never a generic sum — it was the expansion of (x+1)^2, which is why the root simplifies. You did not distribute; you recognised a square.
Taylor approximate (for small terms). In physics and advanced calculations, when one piece is tiny, you use the expansion
This is an approximation, not an identity, and it only works when \varepsilon is close to zero. You will see it often in JEE physics (relativistic corrections, pendulum amplitude corrections) but it is not a licence to split general square roots.
When the identity "looks" to work
Watch out for two near-misses that can fool you into thinking the general rule holds.
Zero sneaks in. \sqrt{1 + 0} = \sqrt{1} = 1, and \sqrt{1} + \sqrt{0} = 1 + 0 = 1. The identity agrees — but only because one operand is zero, which is the trivial case established above. Do not generalise from this.
A hidden perfect square. \sqrt{a^2 + 2ab + b^2} = \sqrt{(a + b)^2} = |a + b|, which equals a + b when both are non-negative. This looks like \sqrt{a^2} + \sqrt{b^2} plus something distributed — but it is not. The reason the root came out cleanly is that the sum inside was already a perfect square. You did not distribute anything; you recognised a square. That is a completely different move.
Recognition drill
Which of the following can you simplify, and which are trapped as written? Go through them in your head before reading the answers.
- \sqrt{25} = 5. Standard — a single perfect square, no split needed.
- \sqrt{x^2 + 9} — trapped. You cannot split this. No factoring, no perfect square. Leave it alone.
- \sqrt{x^2} \cdot \sqrt{9} = |x| \cdot 3 = 3|x|. The multiplication rule applies cleanly.
- \sqrt{x^2 + 6x + 9} = \sqrt{(x + 3)^2} = |x + 3|. Complete-the-square trick works.
- \sqrt{a^2 b^2} = |ab|. A single product under the root — split into |a| \cdot |b| if you like.
- \sqrt{16 + 9} = \sqrt{25} = 5. Not 4 + 3 = 7. You compute the sum inside the root first; you do not distribute.
If any of these tricked you, run the numerical test again. Pick small values, evaluate both sides, and watch where the identity breaks.
The one-line takeaway
The square root does not distribute over addition. It distributes over multiplication. That distinction is a single misconception that costs students marks in every algebra exam and survives decades of classrooms. The defence is cheap: every time you are tempted to write \sqrt{a + b} = \sqrt{a} + \sqrt{b}, plug in a = 9 and b = 16, watch 5 disagree with 7, and put the identity back in the bin where it belongs.