Given a side, find the area — then turn it around: given the area, find the side.
Point 4 of 5 in this lesson: 6.2.4 When √a exists, and why it's never negative
Lay tiles on a square floor. If each side runs 5 tiles long, the floor swallows 5 × 5 = 25 tiles — and we write that 52, read "five squared." That is the easy direction. The whole new idea of this lesson is the reverse trip: a friend tells you a square floor holds 25 tiles and asks how long each side is. Undoing a square is called taking a square root, and by the end of this page you'll read the symbol √ , know exactly what number it points to, and be able to pin down even the "messy" roots like √20 to a tenth.
The steady color habit for this lesson: a side a is teal, the square a², the area is blue, the square-root operation recovers the teal side from the blue area, perfect squares glow blue, and the estimate bounds that trap a messy root are amber.
Why do we say "squared" for the little raised 2? Because of an actual square. Take a side of length a and build a square on it. Slice it into a grid of unit boxes and you get a rows of a boxes each, so the box count — the area — is a × a. We write that product a2 and read it "a squared" or "the square of a." The word and the shape are the same fact.
So squaring a number is exactly "find the area of the square built on it." 32 = 9 because a 3-by-3 square holds 9 unit squares; 72 = 49 because a 7-by-7 square holds 49. The result is an area, and an area is never negative — already a hint about what's coming.
Read each square out loud, then count the area.
62 = 6 × 6 = 36 ("six squared is thirty-six").
102 = 10 × 10 = 100 ("ten squared is one hundred").
A square patio with side 12 ft has area 122 = 144 square feet.
To square a number is to build the square on it and count its tiles: a2 = a × a = the area. Side in, area out.
Step the side a up and down. The grid fills with unit squares, and the readout shows a2 = area.
Now flip the question. A friend says, "my square rug covers 9 square feet — how long is each side?" You are no longer given the side and asked for the area; you are given the area and asked for the side. You need a number that, when squared, gives back 9. That number is 3, because 32 = 9.
Written as an equation, you are solving x2 = 9 — "what number, squared, makes nine?" This undoing of a square is called taking the square root, and it gets its own symbol, the radical sign √ . The little roof over the number is the vinculum; it tells you exactly which area is going under the root. So √9 is read "the square root of nine" and asks: what side gives area nine?
Each square root is a "what side?" question. Find the side whose square is the area.
√16 = 4, because 42 = 16.
√49 = 7, because 72 = 49.
√1 = 1 and √0 = 0 (a side of length 0 makes a square of area 0).
Pick a target area (a perfect square), then step the side until side2 matches the area. When it locks, the root is revealed.
Here is the subtle part. If we only ask "what squares to 9?", there are two answers, because squaring kills minus signs: 32 = 9 and also (−3)2 = (−3) × (−3) = 9. So the equation x2 = 9 has two solutions, x = +3 and x = −3. Both +3 and −3 are "square roots of nine."
But a side length can't be negative, and we want the symbol √ to name one definite number, not two. So mathematicians agreed: √ always means the non-negative one, called the principal square root. Therefore √9 = 3 exactly — never ±3. If you actually want both solutions of x2 = 9, you write the ± yourself: x = ±√9 = ±3.
The number-one mix-up in this whole lesson: writing "√9 = ±3." No. The radical alone is always one non-negative number, so √9 = 3. The ± only appears when you solve an equation like x2 = 9, where two sides could give that area.
Step the candidate side across negatives and positives. Watch both −n and +n land on the same area — but see where √ points.
Because a square root is "the side of a square of area a," two facts fall out for free. First, the number inside the root, the radicand, must satisfy a ≥ 0 — there is no square with a negative area, so there is no real side to find. That means √−4 has no real value: no real number squared gives −4, because squaring any real number lands at 0 or above.
Second, the value that comes out is always ≥ 0 — it's a length. So you will never see a real square root return a negative answer. Put together: √a is defined only when a ≥ 0, and when it is defined, √a ≥ 0. The smallest case is √0 = 0.
| Radicand a | Real value? | Why |
|---|---|---|
| a > 0 | yes, one | a positive side: e.g. √25 = 5 |
| a = 0 | yes, zero | √0 = 0 |
| a < 0 | none | no real square is negative |
Keep the minus sign outside straight in your head. −√25 = −5 is perfectly fine — that's "the negative of √25," and the radicand 25 is still positive. The forbidden thing is a negative under the roof, like √−25, which has no real value.
Slide the radicand a. While a ≥ 0 the root is a real length; push a below zero and the radical lights up "no real value."
Some areas "come out perfectly." The numbers 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, … are exactly the areas of whole-number squares, so their roots are whole numbers. We call them perfect squares. Memorizing the first dozen makes square roots feel instant: √64 = 8, √81 = 9.
But most numbers fall between perfect squares — and their roots are messy (they're irrational, the surprise of the next lesson). You can still pin them down. Take √20. Since 16 < 20 < 25, taking roots gives √16 < √20 < √25, that is 4 < √20 < 5. So the side is between 4 and 5. To refine, test tenths: 4.42 = 19.36 (too small) and 4.52 = 20.25 (too big), so 4.4 < √20 < 4.5. Squeeze again and you get √20 ≈ 4.47.
Estimate √50 to one decimal place.
Bound it. 49 < 50 < 64, so 7 < √50 < 8. It's just barely past 49, so try numbers near 7.
Refine by tenths. 7.12 = 50.41 (a touch too big); 7.02 = 49 (too small). So 7.0 < √50 < 7.1, and since 50 is very close to 50.41, √50 ≈ 7.1 (more precisely 7.07).
Pick a number n. The bar traps √n between consecutive perfect squares; then nudge the tenths slider to tighten the bound.
Squaring builds the square on a side and counts its tiles, so a2 is an area and is never negative; the square root √ walks that trip backward, recovering the side from the area. Two numbers square to 9, namely +3 and −3, but the radical points only at the non-negative principal root, so √9 = 3. A real √a needs a ≥ 0 and returns something ≥ 0. The perfect squares come out whole; everything trapped between them you pin down with amber bounds and refine by tenths.
If a square turns a side into an area, a cube turns an edge into a volume — and we'll undo it with the cube root 3√ . There's one delicious difference: cubes keep the sign, so 3√−8 = −2 is perfectly real.
Work each one out first, then open the answer to check your thinking.
Six questions to lock it in. Tap the answer you think is right.
This lesson serves 8.EE.A.2 — using square-root symbols, evaluating √ of small perfect squares, and knowing that √2 is irrational — with grade-6 area (6.G) as the concrete anchor and 8.NS.A.2 for approximating irrational roots by trapping them between rationals. The #1 misconception is writing "√9 = ±3," conflating the radical (a single non-negative number) with the two solutions of x2 = 9. The antidote: tie the radical to a side length, which can't be negative — the symbol always names the principal (non-negative) root, and the ± appears only when the student solves an equation, where the negative side is also a valid algebraic solution.