Quadratic equations — factoring, completing the square: Visual walkthrough

Before we touch a single symbol, three plain-word promises:

$x$ is just an unknown length — think of it as the side of a square we haven't measured yet.
$x^{2}$ means $x \times x$ — the area of that square (a side times a side).
$b x$ means " $b$ copies of a strip that is $x$ long" — a rectangle with one side $x$ and the other side $b$ .

Everything below is a game of cutting rectangles and re-gluing them into a square. That is the entire trick.

Step 1 — Read the equation as areas

WHAT. Take the plainest quadratic, $x^{2} + b x = c^{'}$ (I moved the constant to the right; $c^{'}$ is just "whatever number is left over"). Here $b > 0$ for now.

WHY. We cannot "complete a square" until we see the pieces as shapes. An algebra equation is invisible; a picture of tiles is not. So the first move is always: turn each term into a region whose area equals that term.

PICTURE. Look at Figure s01.

The big teal square has side $x$ , so its area is $x \cdot x = x^{2}$ . Each symbol is doing its job right there: the two sides are both labelled $x$ , and their product fills the square.
The orange rectangle is $x$ tall and $b$ wide, so its area is $b \cdot x = b x$ . The side labelled $b$ is the width; the side labelled $x$ is the height.

Together the two tiles have area $x^{2} + b x$ , which the equation says equals $c^{'}$ .

Step 2 — Slice the strip in half and move it

WHAT. Cut the orange $b$ -wide rectangle down the middle into two thinner rectangles, each of width $\frac{b}{2}$ . Swing one of them from the side of the square to the bottom.

WHY. A square is symmetric — its extra pieces should hug it evenly on two sides, not pile up on one. Splitting $b x$ into two equal halves $\frac{b}{2} x + \frac{b}{2} x$ lets us stick one half on the right of the square and one half underneath. This starts shaping the whole thing into a bigger square.

PICTURE. In Figure s02:

The width $b$ became two widths of $\frac{b}{2}$ (the label on each thin strip).
One strip stays on the right edge of the teal square; the other rotates down to the bottom edge. Each strip is still $x$ long in one direction and $\frac{b}{2}$ in the other, so each has area $\frac{b}{2} \cdot x$ . Two of them: $2 \cdot \frac{b}{2} x = b x$ — no area was lost, we only rearranged.

Notice the shape now: an $x$ -by- $x$ square with two $\frac{b}{2}$ -thick flaps forming an L. There is an obvious empty corner.

Step 3 — Fill the missing corner

WHAT. The L-shape is missing exactly one little square at the bottom-right corner. Its two sides are both $\frac{b}{2}$ (the flap thicknesses), so its area is $(\frac{b}{2})^{2}$ . Add that tile.

WHY. This is the whole point of "completing the square." Once we drop in that corner tile, the L-shape closes up into one big solid square. Its side is $x + \frac{b}{2}$ (the original $x$ plus one flap thickness). We literally complete the square by supplying its one missing piece.

PICTURE. Figure s03 shows the plum corner tile snapping in.

Each side of the completed square reads $x + \frac{b}{2}$ .
The whole area is now $(x + \frac{b}{2})^{2}$ .
The added plum tile has area $(\frac{b}{2})^{2} = \frac{b ^{2}}{4}$ — that is the number we introduced.

Step 4 — Keep the equation honest (add to BOTH sides)

WHAT. We added area $\frac{b ^{2}}{4}$ on the picture's left, so we must add the same number to the right side of the equation.

WHY. The equation $x^{2} + b x = c^{'}$ is a balance scale. If you drop a weight $\frac{b ^{2}}{4}$ on the left pan, the scale tips unless you drop the same weight on the right pan. This is exactly the parent note's Mistake 3 — adding to one side only breaks the truth of the equation.

PICTURE. Figure s04 is the balance: $L-shape x^{2} + b x + corner tile \frac{b ^{2}}{4} = c^{'} + \frac{b ^{2}}{4} .$

Left pan: the completed square $(x + \frac{b}{2})^{2}$ .
Right pan: the old number $c^{'}$ plus the same $\frac{b ^{2}}{4}$ .

So the left side collapses to a single squared term: $(x + \frac{b}{2})^{2} = c^{'} + \frac{b ^{2}}{4} .$

Step 5 — Un-square: take the root (BOTH signs)

WHAT. The left side is a square. Undo it by taking a square root. But a square root of a positive number has two answers, one $+$ and one $-$ .

WHY. Ask: "what side length, squared, gives this area?" That question is answered by . But squaring forgets signs: $(+ 3)^{2}$ and $(- 3)^{2}$ both give $9$ . So both candidates must be kept, or we'd lose a root. That is where the $\pm$ comes from — it is not decoration, it is honesty about two possible sides.

PICTURE. Figure s05 shows the number line: the value $x + \frac{b}{2}$ can sit the same distance $c^{'} + \frac{b ^{2}}{4}$ to the right ( $+$ ) or to the left ( $-$ ) of zero. $x + \frac{b}{2} = \pm c^{'} + \frac{b ^{2}}{4} .$ Then slide $\frac{b}{2}$ back over: $x = - \frac{b}{2} \pm c^{'} + \frac{b ^{2}}{4} .$

$- \frac{b}{2}$ is the centre of the two roots (the axis of symmetry — see 2.3.4-Parabolas-and-vertex-form).
$\pm \dots$ is the equal distance each root sits from that centre.

Step 6 — Worked run: $x^{2} + 6 x + 5 = 0$

WHAT. Put real numbers through the exact same tiles. Here $b = 6$ , and moving the $5$ over gives $c^{'} = - 5$ .

WHY. To trust the picture, watch it produce the parent note's Example 3 answer.

PICTURE. Figure s06 uses tiles of real size.

Strip width $b = 6$ , so each half-strip is $\frac{6}{2} = 3$ wide.
Missing corner tile side $3$ , area $3^{2} = 9$ .
Add $9$ to both sides: $x^{2} + 6 x + 9 = - 5 + 9 = 4$ , i.e. $(x + 3)^{2} = 4$ .
Un-square: $x + 3 = \pm 2$ , so $x = - 3 + 2 = - 1$ or $x = - 3 - 2 = - 5$ .

Both match Example 3 exactly. ✓

Step 7 — The degenerate & sign cases (never skip these)

WHAT. Three scenarios the tile game must still handle.

WHY. A method you only trust for "nice" numbers is not a method. Check the edges.

Right side zero ( $c^{'} + \frac{b ^{2}}{4} = 0$ ): the square has area $0$ , so its side is $0$ . Then $x + \frac{b}{2} = 0$ and there is one repeated root $x = - \frac{b}{2}$ . The parabola just kisses the axis.
Right side negative ( $c^{'} + \frac{b ^{2}}{4} < 0$ ): no real square has negative area, so $negative$ is not a real number — no real roots. The parabola floats entirely above (or below) the axis. This is exactly the discriminant $b^{2} - 4 a c < 0$ showing up as "impossible tile area."
$b < 0$ (e.g. $x^{2} - 4 x = \dots$ ): the strip points the other way, but we still halve it: $\frac{b}{2} = - 2$ , and the corner tile area $(\frac{b}{2})^{2} = (- 2)^{2} = 4$ is still positive (a square of area is never negative). So the recipe is unchanged — only the centre $- \frac{b}{2}$ moves to the positive side. This is why Example 4's $2 x^{2} - 8 x + 3$ centres at $x = 2$ .

PICTURE. Figure s07 stacks all three: a zero-area point (one root), an "impossible" negative-area ghost (no roots), and a flipped-strip case (still works).

The one-picture summary

Figure s08 compresses the whole journey onto one canvas: square $\to$ split strip $\to$ add corner $\to$ balance both sides $\to$ root with $\pm$ . Follow the arrows left to right and you have re-derived the quadratic formula $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ without memorising a thing — the $- b$ is the centre, the is the corner-completed area, the $\pm$ is the two sides, the $2 a$ is dividing by the leading coefficient before you start.

Recall Feynman retelling (plain words)

I had an unknown square of side $x$ , plus a long rectangle stuck to it — that rectangle was the " $b x$ " part. To make everything into one clean square, I cut the rectangle in half and laid the two halves along two edges of my square, making an L. An L is almost a square — it just needs one little corner filled. That corner is a tiny square whose side is half the rectangle's width, so its area is $(b /2)^{2}$ . I dropped that tile in — but to keep the equation fair I dropped the same amount on the other side of the equals sign. Now my left side is a genuine big square of side $x + b /2$ . To find $x$ I ask "what side gives this area?" — that's a square root — and I remember a side could point either way, so I write $\pm$ . Finally I slide the $b /2$ back and read off both answers. Same story every time: even a repeated root (corner area zero) or an impossible one (would-be negative area) just falls out of what the tiles allow.

Recall Quick self-checks

Why do we halve $b$ before squaring? ::: Because the strip $b x$ splits into two equal flaps of width $b /2$ ; the missing corner tile has that same side. Where does the $\pm$ come from? ::: Un-squaring a square: two side lengths (one $+$ , one $-$ ) give the same area, so both roots survive. What does a negative right-hand side mean geometrically? ::: A square with negative area is impossible, so there are no real roots (discriminant $< 0$ ). Why add the corner tile to both sides? ::: The equation is a balance; adding area to one pan only would break the equality.

Related paths: 3.2.1-Polynomials-factoring-techniques, 4.1.2-Solving-systems-graphically, 2.1.15-Linear-equations-and-inequalities.