Quadratic formula — derivation by completing the square: Visual walkthrough

Before we start, one promise: I will not write a symbol you have not met. Let me name the four characters.

We attack this in pictures. Squaring a length gives an area — that is the bridge between algebra and geometry we will ride the whole way.

Step 1 — Make the big square have side 1·x

WHAT. We start from $a x^{2} + b x + c = 0$ and divide every term by $a$ : $a clean square x^{2} + new width \frac{b}{a} x + new constant \frac{c}{a} = 0$

WHY. To draw $x^{2}$ as a square whose side is exactly $x$ , its coefficient must be $1$ . Dividing by $a$ is legal precisely because $a \neq = 0$ — you may never divide by zero, and this is the one place the " $a \neq = 0$ " rule earns its keep.

PICTURE. Look at the figure. The red square has side $x$ , so its area is $x^{2}$ . Sitting beside it is a rectangle of height $x$ and width $\frac{b}{a}$ — area $\frac{b}{a} x$ . Everything is now expressed as areas of tiles with side or height $x$ .

Figure — Quadratic formula — derivation by completing the square

Step 2 — Push the constant off the board

WHAT. Move the constant term to the right: $x^{2} + \frac{b}{a} x = - \frac{c}{a}$

WHY. We want the left side to become one perfect square shape. To do that we must first clear the board of the loose constant tile, leaving only the pieces we can rearrange into a square: the $x^{2}$ tile and the $\frac{b}{a} x$ strip. The right side, $- \frac{c}{a}$ , is just "the total area we are aiming for."

PICTURE. The left panel keeps the square + strip. The right panel is a single quantity $- \frac{c}{a}$ we must eventually match.

Step 3 — Split the strip in half and swing it around

WHAT. Take the strip of area $\frac{b}{a} x$ and cut it into two equal halves, each of width $\frac{b}{2 a}$ and area $\frac{b}{2 a} x$ . Glue one half to the right edge of the square and one half to the bottom.

WHY. A single long strip on one side can never close up into a square. But split symmetrically onto two adjacent sides, the pieces frame an almost-complete square — an L-shape with exactly one corner missing. The number $\frac{b}{2 a}$ (the half-width) is going to become the $\frac{b}{2 a}$ inside the final formula: watch it appear here first.

PICTURE. The two red half-strips hug two sides of the black square. Notice the empty little corner at the far vertex — that is the piece we are missing.

Step 4 — Complete the square (fill the missing corner)

WHAT. The missing corner is a small square of side $\frac{b}{2 a}$ , so its area is $(\frac{b}{2 a})^{2} = \frac{b ^{2}}{4 a ^{2}}$ . We add it to both sides: $x^{2} + \frac{b}{a} x + corner tile (\frac{b}{2 a})^{2} = - \frac{c}{a} + same tile, added (\frac{b}{2 a})^{2}$

WHY. Adding only to the left would change the equation — cheating. Adding the identical tile to both sides keeps the balance perfect while finishing the square. This is the move; everything after is bookkeeping.

The whole left side is now the area of one big square whose side is $x + \frac{b}{2 a}$ : $(x + \frac{b}{2 a})^{2}$ Here $x$ is the original side and $\frac{b}{2 a}$ is the half-strip width we bolted on — together they form the full side length.

PICTURE. The little red corner drops into place; the L-shape closes into a full square of side $x + \frac{b}{2 a}$ .

Step 5 — Tidy the right side into one fraction

WHAT. Combine the right-hand side over a common denominator $4 a^{2}$ : $- \frac{c}{a} + \frac{b ^{2}}{4 a ^{2}} = - \frac{4 a c}{4 a ^{2}} + \frac{b ^{2}}{4 a ^{2}} = \frac{b ^{2} - 4 a c the discriminant Δ}{4 a ^{2}}$

So the picture-equation reads: $(x + \frac{b}{2 a})^{2} = \frac{b ^{2} - 4 a c}{4 a ^{2}}$

WHY. To undo a square we take a square root, and a square root is cleanest over a single fraction. Merging the two right-side pieces prepares that. The top of this fraction, $b^{2} - 4 a c$ , is so important it gets a name — the discriminant $Δ$ — because it decides everything about the answer (Step 7).

PICTURE. Left area (the completed square) equals right area (one combined rectangle). Equal areas, drawn side by side.

Step 6 — Take the square root, keep BOTH signs

WHAT. Undo the squaring: $x + \frac{b}{2 a} = \pm \frac{b ^{2} - 4 a c}{2 a}$ then move $\frac{b}{2 a}$ across and combine: $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$

WHY the $\pm$ . A square erases the sign of its input: $(+ 3)^{2} = (- 3)^{2} = 9$ . So when we run it backward, two side-lengths could have produced that area — a positive one and a negative one. Honesty forces us to keep both, and that is where the two roots of a quadratic are born.

Term-by-term, right where each lives: $x = \frac{- b swing \frac{b}{2 a} across \pm b ^{2} - 4 a c side from Step 5}{from 2 a under the root 2 a}$

PICTURE. One target area splits into two possible signed side-lengths — the two crossing points where the curve meets zero.

Step 7 — The three faces of $Δ$ (all cases)

The root hides inside $Δ$ , so the sign of $Δ = b^{2} - 4 a c$ controls the whole story. We must cover every possibility.

WHAT & WHY (case by case):

$Δ > 0$ — the area under the root is positive, its square root is a real number, and the $\pm$ gives two different real answers. Curve crosses the axis twice.
$Δ = 0$ — the area is zero, $0 = 0$ , so $+ 0$ and $- 0$ collapse to one repeated root. Curve just touches the axis.
$Δ < 0$ — you cannot make a real square with negative area. There are no real roots; the answers live in the complex numbers using $- k = i k$ . Curve misses the axis entirely.

PICTURE. Three parabolas, one per case, against the same axis — cross, touch, miss.

The one-picture summary

Read left to right: bare square + strip $\to$ split + missing corner $\to$ completed square = combined area $\to$ square-root splits into the two roots. That is the entire derivation in one strip.

Recall Feynman retelling — say it like a story

We had a lumpy shape: one square tile ( $x^{2}$ ) and a long strip ( $\frac{b}{a} x$ ). A lumpy shape is hard to measure. So we sliced the strip down the middle and wrapped the two halves around two sides of the tile — now it almost looks like a bigger square, with just one tiny corner hole. We filled that hole (adding the same tile to the other side of the scale so nothing tips). Now the left is a genuine square of side $x + \frac{b}{2 a}$ , and its area equals a neat fraction $\frac{b ^{2} - 4 a c}{4 a ^{2}}$ . To find the side from the area we take a square root — and because a square forgets whether its side was plus or minus, we keep both signs. Sliding the $\frac{b}{2 a}$ back across gives $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ . Whether the thing under the root is positive (two answers), zero (one), or negative (none real, two complex) — that is the whole life of a quadratic, drawn as tiles.

Recall

What does splitting the middle strip into two halves of width $\frac{b}{2 a}$ eventually become in the final formula? ::: The $\frac{b}{2 a}$ we swing across the equals sign, giving the $- b$ over $2 a$ . Why does the $\pm$ appear? ::: Squaring destroys sign information, so a positive and a negative side both give the same area — both must be kept as roots. What geometric event corresponds to $Δ = 0$ ? ::: The missing-corner-completed square has zero area on the right, so the two roots merge into one — the parabola just touches the axis. Why is $a \neq = 0$ essential? ::: Dividing by $a$ (Step 1) and the $2 a$ denominator both require $a \neq = 0$ ; if $a = 0$ the equation is linear, not quadratic.

Connections

Parent: full algebraic derivation
Completing the square technique
Perfect square trinomials
Discriminant and nature of roots
Complex numbers and quadratic equations
Parabola and its vertex form
Factoring quadratics
Vieta's formulas for sum and product of roots