Introduction
The moment you write this on a test, your points vanish.
$$\sqrt{9 - 4} = \sqrt{9} - \sqrt{4} = 3 - 2 = 1$$It looks right, but it's wrong. In reality, \(\sqrt{9-4} = \sqrt{5} \approx 2.24\).
This kind of mistake comes from believing you can just "peel the root off" the inside of the parentheses. But a root splits only across multiplication and division, and it does not split across addition and subtraction. In this post we'll chase down exactly why, all the way through, using a single method: squaring and comparing. The goal isn't to memorize, but to make it visible.
Start from the definition of a root
First, let's pin down again what \(\sqrt{a}\) actually means.
\(\sqrt{a}\) is "the non-negative number that, when squared, gives \(a\)." (Here we deal only with the case \(a \ge 0\).)
So the single most important property of \(\sqrt{a}\) is just this:
$$ \left(\sqrt{a}\right)^2 = a $$From now on, every judgment rests on this one line. "When you square an expression, does exactly what was inside come back out?" — if so, that expression really is the root of that number. With this single criterion, we'll test multiplication, addition, and subtraction in turn.
What works — multiplication and division
To check whether \(\sqrt{a}\cdot\sqrt{b}\) equals \(\sqrt{ab}\), we just square \(\sqrt{a}\cdot\sqrt{b}\) and see whether \(ab\) comes out.
$$ \left(\sqrt{a}\cdot\sqrt{b}\right)^2 = \left(\sqrt{a}\right)^2 \cdot \left(\sqrt{b}\right)^2 = a \cdot b $$Squaring it gave exactly \(ab\). Not a single leftover term. So \(\sqrt{a}\cdot\sqrt{b}\) really is the root of \(ab\).
$$ \sqrt{ab} = \sqrt{a}\cdot\sqrt{b} $$Division holds for the same reason.
$$ \left(\frac{\sqrt{a}}{\sqrt{b}}\right)^2 = \frac{a}{b} \quad\Longrightarrow\quad \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}} \qquad (b > 0) $$These two rules always hold. The key was that "when you square, only what was inside is left, cleanly."
What doesn't work — addition and subtraction
So why does addition fail? Let's square \(\sqrt{a} + \sqrt{b}\) the same way. What's different from the multiplication case becomes clear right here.
Apply the special product \((p+q)^2 = p^2 + 2pq + q^2\) directly with \(p=\sqrt{a},\ q=\sqrt{b}\).
$$ \left(\sqrt{a} + \sqrt{b}\right)^2 = \left(\sqrt{a}\right)^2 + 2\sqrt{a}\sqrt{b} + \left(\sqrt{b}\right)^2 = a + 2\sqrt{ab} + b $$Looking at the result, besides \(a + b\) there's an extra term, \(2\sqrt{ab}\), tacked on. What we wanted was for squaring to give exactly \(a+b\) — but a leftover appeared.
$$ \left(\sqrt{a} + \sqrt{b}\right)^2 = a + 2\sqrt{ab} + b \;\neq\; a + b $$If \(a, b\) are both positive, then \(2\sqrt{ab} > 0\), so this extra term never disappears. Therefore \(\sqrt{a} + \sqrt{b}\) is not the root of \(a+b\).
$$ \sqrt{a+b} \neq \sqrt{a} + \sqrt{b} \qquad \text{(in general)} $$Subtraction breaks down the same way. Squaring \(\sqrt{a} - \sqrt{b}\) gives
$$ \left(\sqrt{a} - \sqrt{b}\right)^2 = a - 2\sqrt{ab} + b \;\neq\; a - b $$Again, because of the extra term \(-2\sqrt{ab}\), \(a-b\) doesn't come out cleanly.
$$ \sqrt{a-b} \neq \sqrt{a} - \sqrt{b} \qquad (a > b > 0) $$Checking with numbers
Let's go back to the original \(\sqrt{9-4}\).
| Expression | Computation |
|---|---|
| \(\sqrt{9-4}\) | \(\sqrt{5} \approx 2.24\) |
| \(\sqrt{9} - \sqrt{4}\) | \(3 - 2 = 1\) |
| Difference | \(2.24 - 1 = 1.24 \neq 0\) |
The extra term confirms it too. The discrepancy is driven by \(2\sqrt{ab} = 2\sqrt{9\cdot 4} = 2\cdot 6 = 12\). (That's the term produced at the squaring step.)
To summarize:
A root splits only across × and ÷. Across + and −, it cannot be split. The reason is a single one — squaring an addition or subtraction produces the extra term \(2\sqrt{ab}\).
Try it yourself
Move the A and B sliders to compare three expressions in real time.
- \(\sqrt{A\times B}\) vs. \(\sqrt{A}\times\sqrt{B}\) → always match (✓)
- \(\sqrt{A+B}\) vs. \(\sqrt{A}+\sqrt{B}\) → don't match (✗)
- \(\sqrt{A-B}\) vs. \(\sqrt{A}-\sqrt{B}\) → don't match (✗)
With the defaults A=9, B=4, you can see right away that \(\sqrt{9-4} = \sqrt{5} \approx 2.24\) differs from \(3-2=1\). Check whether, no matter which A and B you plug in, only the multiplication row stays ✓.
Key takeaways
| Operation | Splittable? | When you square it |
|---|---|---|
| Multiplication \(\sqrt{ab}\) | ✓ yes | \((\sqrt{a}\sqrt{b})^2 = ab\) — no extra term |
| Division \(\sqrt{a/b}\) | ✓ yes | \((\sqrt{a}/\sqrt{b})^2 = a/b\) — no extra term |
| Addition \(\sqrt{a+b}\) | ✗ no | \((\sqrt{a}+\sqrt{b})^2 = a + 2\sqrt{ab} + b\) |
| Subtraction \(\sqrt{a-b}\) | ✗ no | \((\sqrt{a}-\sqrt{b})^2 = a - 2\sqrt{ab} + b\) |
The criterion is always the same. "When you square it, does the inside come back out with no leftovers?" Multiplication and division do; addition and subtraction don't, because \(2\sqrt{ab}\) is left behind.
Studying this with AI
When you ask "why doesn't this work," AI will usually generate a counterexample (numbers) right away. For instance, if you ask "show me, by squaring, why √(a+b)=√a+√b is wrong," you can get the step-by-step expansion above. That said, it's important to cross-check the numbers AI gives you by plugging values in yourself, just like the interactive above.
Wrapping up
The root of this mistake is failing to check "how far a rule actually applies."
- Splitting (distributing) a root applies only to × and ÷.
- Across + and −, don't split — compute the inside of the parentheses first.
- When in doubt, "square and compare" — if the extra term \(2\sqrt{ab}\) shows up, it's a sign that splitting is not allowed.
Related reading — How do you track the sign of a one-sided limit? · What on earth are trigonometric functions for? · Why does the arctangent appear? · Computational check: Wolfram Alpha