Where a squared term comes from, what counts as a quadratic, and the first way to solve one.
Picture a square vegetable garden. Someone tells you it covers exactly 36 square metres, and you want to know how long each side is. In Stage 10 every unknown you chased grew in a straight line — "how far did I walk," "how much did it cost." But the area of a square does not come from adding the side to itself; it comes from multiplying the side by itself. The unknown shows up squared. That little change — an x2 instead of an x — is the entire subject of this chapter, and this lesson is the first time we meet it.
By the end you will be able to say what a quadratic equation is, write any of them in the tidy standard form ax2 + bx + c = 0, read off its three numbers, check whether a value is a root, and solve the simplest quadratics by taking a square root. Throughout the chapter we keep the same colors: a / the squared term x2 in blue, b / the linear term x in teal, c / the constant in amber, and the roots — the values of x that solve the equation — in violet.
Compare two ways a square can grow. Its perimeter — the distance all the way around — is 4 × side. That grows in lockstep with the side: a fence around a side-3 square needs 12 m of fencing; double the side to 6 and you need 24 m of fencing, exactly twice as much. Perimeter is a linear quantity. Its area behaves completely differently, because area is side × side = side2.
Watch what doubling does to area. A side-3 square has area 3 · 3 = 9. A side-6 square has area 6 · 6 = 36. The side doubled, but the area went up four times, not two. That is no accident: if you double a side s, the new area is (2s)2 = 4s2 — four copies of the old area. Triple the side and the area becomes nine times bigger. Area races ahead of length because both factors grow at once.
Now run the whole thing backwards. Instead of being handed the side and asked for the area, you are handed the area and asked for the side. That is the garden problem: s2 = 36, find s. The squaring has to be undone, and the answer s = 6 is the value that, multiplied by itself, rebuilds 36. That reversal — "the unknown is squared, and I have to dig it back out" — is the seed of every quadratic equation in this chapter.
Anytime two quantities multiply to make the thing you care about, you get a square. Areas (side × side). The height of a thrown or falling object after t seconds (gravity makes distance depend on t × t). Anything built from a rate times a rate. Whenever the answer depends on a quantity multiplied by itself, the equation that finds that quantity is quadratic.
Drag the side from 1 up to 8 and watch how fast the number of unit squares climbs.
Here is the definition in plain words: a quadratic equation has one unknown, and the highest power of that unknown is exactly 2. Once you tidy it up — gather every term on one side so the other side is 0 — it always reads
ax2 + bx + c = 0, with a ≠ 0.
The word "quadratic" comes from the Latin quadratus, meaning square — a fitting name for the equation that grew out of a square garden. The condition a ≠ 0 matters more than it looks. If a were 0, the ax2 term would vanish entirely and you would be left with bx + c = 0 — an ordinary linear equation from Stage 10, not a quadratic at all. The x2 term is what makes a quadratic a quadratic, so its coefficient is not allowed to disappear.
x2 − 5x + 6 = 0 (the full template). 2x2 = 8 (highest power 2; tidy it to 2x2 − 8 = 0). 3x2 − 12 = 0 (here b = 0, and that is fine — only a must be nonzero). (x + 1)2 = 9 (multiply it out and an x2 appears).
2x + 1 = 0 — the highest power is 1, so it is linear. 1/x = 5 — here x sits in a denominator, which is not a polynomial at all. And anything with an x3 that does not cancel is degree 3, higher than quadratic.
Choose the three coefficients and watch the equation assemble. Set a to 0 and see what breaks.
Once an equation is in standard form, the three numbers a, b, c are just sitting there waiting to be read off — in order: the number on x2, the number on x, and the lonely constant. The catch is that most equations do not arrive in standard form. You have to tidy first, then read. And the single most common mistake happens during the tidying: forgetting the sign change that comes from moving a term across the equals sign.
3x2 = 2x − 5. Move everything to the left: subtract 2x and add 5. You get 3x2 − 2x + 5 = 0, so a = 3, b = −2, c = 5. Notice the −2: the +2x became −2x when it crossed over.
x2 = 9. There is no x term, but every quadratic has one — it is just hiding with coefficient 0. Write x2 + 0x − 9 = 0, so a = 1, b = 0, c = −9.
5 − x2 = 0. Put the squared term first: −x2 + 5 = 0. We prefer a to be positive, so multiply every term by −1: x2 − 5 = 0, giving a = 1, b = 0, c = −5. If you forget to flip the 5 you will wrongly write c = +5. Multiply through, every term, or not at all.
Step through messy equations. Each one is tidied to standard form, then a, b, c are read off with their signs.
A root — also called a solution — is a value of x that makes the equation actually balance. Plug it in, work out both sides, and they agree; equivalently, ax2 + bx + c comes out to exactly 0. There is no magic to checking a candidate: you substitute the number and see whether it works.
Try x = 2: 22 − 5·2 + 6 = 4 − 10 + 6 = 0 ✓ — a root.
Try x = 3: 32 − 5·3 + 6 = 9 − 15 + 6 = 0 ✓ — also a root.
Try x = 1: 12 − 5·1 + 6 = 1 − 5 + 6 = 2 — this is 2, not 0, so x = 1 is not a root.
That example shows something important: a quadratic usually has two roots, not one. (This makes sense — squaring loses the sign, so two different inputs can land on the same output.) "Solving" a quadratic means finding all of its roots, not just the first one you stumble on.
Slide x through the whole numbers and watch the substitution. Two values make it land on 0 — find them.
Some quadratics wear their answer on their sleeve. Whenever an equation can be arranged to look like x2 = k — a squared thing equal to a number — you can solve it directly by undoing the square: take the square root of both sides. The one thing you must never forget is that two numbers square to give the same positive k, one positive and one negative. So the answer is x = ±√k, with that ± doing real work.
x2 = 49 → x = ±√49 = ±7. Check: 72 = 49 and (−7)2 = 49. Both work.
x2 = 10 → x = ±√10 ≈ ±3.16. The root is not a whole number, and that is perfectly fine — leave it as √10 or round it.
(x − 3)2 = 16 → take the root of both sides: x − 3 = ±4. So x − 3 = 4 gives x = 7, and x − 3 = −4 gives x = −1.
x2 = −4 → no real solution. No real number, positive or negative, squares to a negative — squaring always lands on 0 or above.
Writing x2 = 25 → x = 5 and stopping. You dropped the negative root! The full answer is x = ±5. The only time you keep just the positive root is when the problem itself forbids the negative one — like the garden, where a side length can't be negative, so s2 = 36 gives s = 6 only. Drop the negative root for a reason, never by accident.
Switch between the two shapes. In each, watch the two violet roots sit symmetrically on the number line.
An unknown gets squared whenever it is multiplied by itself — the way a square's area is side × side — and reversing that squaring is what a quadratic equation asks of us. Tidied onto one side, every quadratic reads ax2 + bx + c = 0 with a ≠ 0, and its three signed coefficients are read off only after you reach standard form. A root is any value of x that makes the left side equal 0, found or checked by plain substitution, and a quadratic usually has two of them. When the equation collapses to x2 = k, you solve it on the spot by taking a square root — remembering the ±, and remembering that a negative k has no real root at all.
Taking square roots only works when the quadratic is already a perfect square sitting equal to a number. In the next lesson we learn completing the square: a clever way to force any quadratic into that friendly (x + m)2 = k shape, so the square-root trick you just learned can finish the job every time.
Work each one out first, then open the answer to check your thinking.
Tap the answer you think is right.
This lesson aligns with the introduction of quadratic equations in a typical US Algebra I–II sequence: writing and interpreting standard form (A-SSE.A, A-CED.A) and solving by the square-root method (A-REI.B.4b). The misconceptions to watch for are three. First, that x2 equals 2x — it does not; x2 means x · x, so x = 3 gives 9 while 2x gives 6. Second, the dropped negative root: students reflexively write √k as one number when x2 = k has two solutions, ±√k. Third, sign errors when moving terms to reach standard form. The square-garden model is chosen because area-from-side is the most physical reason an unknown ever gets squared, and it justifies keeping only the positive root in context (a length can't be negative) without making "drop the negative" a blind habit. The square-root method here is deliberately the gentle on-ramp to completing the square in lesson 11.2.