See a quadratic turn into a perfect square, one tile at a time

How to use this simulation

1. Before touching anything, look at the tiles: the simulation opens on x² + 6x + 5. The big blue square is x², the strip beside it is the 6x part, and the green block is the constant 5. Press Play and the strip is cut in half, one half moves under the square, and a missing corner appears.
2. Drag the “Step” slider from 0 to 4 to move through the stages by hand. Step 1 splits the strip, step 2 fills the dashed corner of area 9, step 3 shows the finished square (x + 3)² and step 4 reads off the roots −5, −1 on the small parabola.
3. Change b. Make it negative and the strip turns red: negative tiles. The corner stays positive, because a negative number squared is positive. Try an odd b as well and watch the half-strips become fractions.
4. Change c. Watch the two small squares under the parabola. When the accent square (b/2)² is bigger than the constant, the parabola crosses the x-axis. When it is smaller, the curve floats above the axis and the simulation says “No real roots”.
5. Tick “Let the x² coefficient a differ from 1” to bring in a leading coefficient. The tiles are then drawn after dividing through by a, which is exactly the first step you must write in an exam.
6. Leave “Show the working” on. Under the picture, every line of the solution appears with your own numbers: the half, the corner, the square, the vertex and the roots.

Al-Khwarizmi really did complete a square

Around the year 820, in Baghdad, the scholar al-Khwarizmi wrote the book whose title gave us the word “algebra”. One of its first problems is a square and 10 roots equal 39. In modern symbols that is x² + 10x = 39. He had no letters and no negative numbers, so he solved it with a drawing.

Draw a square of side x. Hang a strip of width 5 on one side and another on the bottom, because half of 10 is 5. Together the three pieces have area 39, but a corner is missing: a small square of side 5, area 25. Add it to the picture and you must add it to the 39 too. The big square now has area 64, so its side is 8.

The side is x + 5, so x = 8 − 5 = 3. Check it: 3² + 10×3 = 39. The negative answer −13 was ignored in 820, since a length cannot be negative, but today we keep it. This page is that drawing, made to move.

Perfect squares, from zero

A perfect square is something multiplied by itself. Numbers first: 25 is 5 × 5. In algebra, (x + 3) multiplied by itself is x² + 3x + 3x + 9 = x² + 6x + 9. Two middle terms of 3x merge into one 6x.

Look at the answer and spot the pattern. The middle coefficient is double the number in the bracket, and the last term is the number in the bracket squared. Go the other way and you have the recipe for completing the square: half the middle coefficient, squared, is the constant that makes a perfect square.

Words you will meet

Keep this table at hand. The same ideas go by different names in different textbooks.

The tile picture: what you see in the simulation

Think of x as an unknown length. The square of side x has area x², the blue tile. A strip x long and 1 wide has area x. Putting b of those strips side by side makes one strip of area bx, the second colour. The constant c is a block of c unit squares, green.

A tile is positive when it is solid and negative when it is red. If b is negative, the whole strip is red. That is why a red strip is still a strip with the same length: the colour only records the sign.

The algebra, step by step

Write the recipe as five lines. Use x² + 6x + 5 = 0 as the example, the simulation’s own starting point.

Where the quadratic formula comes from

The famous formula is not magic. It is completing the square done once, with letters, so you never have to do it again. Follow the six lines below; each is the previous one with one move.

Try it in the simulation

Five experiments. Each takes under a minute and each teaches something an exam will ask.

• Reset, then slide c to 9. The corner and the constant become the same size, Δ = 0, and there is one repeated root, −3. The parabola just touches the x-axis.
• Push c up to 13. The constant now beats the corner. Δ = −16, the curve floats above the axis, and the roots readout says “No real roots”.
• Set b to −6. The strip turns red and the bracket becomes (x − 3)². The vertex moves to the right side of the axis, to x = 3.
• Set b to 0. There is no strip to cut, no corner, and x² + c = 0 is already a square. This is the easy special case.
• Tick the a toggle and set a = 2. Watch the tiles redraw after dividing through by 2, while the readings show the original quadratic. The roots are unchanged by the division, but the vertex height is multiplied back by a.

Solved problems

Every number below is worked out by the same code that drives the simulation, so you can check each line on the screen.

Common mistakes

These are the places marks leak away. Each has a quick fix.

• Forgetting to divide by a. If a ≠ 1 divide the whole equation first, constant included, or the corner will be wrong.
• Taking the wrong sign. The bracket is x + b/2, not x − b/2. A negative b gives (x − something)², and the root is x = +something ± the spread.
• Adding the corner but not subtracting it. If you add (b/2)² on one side of an equation you must add it on the other side too.
• Squaring b instead of b/2. The corner is (b/2)², not b², and not b/2 squared wrongly as b²/2.
• Dropping the ± when taking the square root. Both signs give a root; losing one loses half the answer.
• Declaring a root when the right side is negative. If the square equals a negative number, there are no real roots.

Why this matters beyond one equation

The vertex form is the practical payoff. x² + 6x + 5 has its vertex at (−3, −4) just by reading (x + 3)² − 4. With a positive coefficient the parabola opens up and this is the minimum; with a negative one it opens down and it is the maximum.

Exam corner

Examiners in CBSE Class 10, GCSE and A level all love this method, because it is the only one that works for every quadratic and shows your reasoning. Marks go to the layout.

• Write “divide by a” as its own line when a ≠ 1.
• Write the corner (b/2a)² explicitly before using it.
• Show the line “= (x + …)² − …”. That line carries the method mark.
• Leave a root as a surd unless told to give decimals, and say to how many places if you do.
• If asked for the nature of roots, quote Δ before solving.

Quick revision

Pin these to the wall before the exam.

• Perfect square: (x + h)² = x² + 2hx + h².
• To complete x² + bx, add (b/2)².
• x² + bx + c = (x + b/2)² − (b/2)² + c.
• Vertex of ax² + bx + c: (−b/2a, c − b²/4a).
• Roots: x = (−b ± √(b² − 4ac)) / 2a.
• Δ > 0 two real roots, Δ = 0 one repeated root, Δ < 0 no real roots.
• A negative b gives a red strip, a negative bracket and the same corner.

Frequently asked questions