Discriminant — nature of roots (real - equal - complex): Visual walkthrough

This page rebuilds the discriminant one picture at a time. We start with a curve, watch it slide, and discover the number $b^{2} - 4 a c$ instead of being handed it. Every symbol is earned before it appears.

Step 1 — What is a quadratic, as a shape?

Each symbol, right where it lives:

$y = how steep / which way the U opens a x^{2} + tilts & slides the U b x + height where x = 0 c$

We require $a \neq = 0$ . If $a = 0$ the $x^{2}$ term vanishes and the U flattens into a straight line — a different creature entirely (we return to this in Step 8).

Step 2 — Why we cannot just "read off" the roots

The goal of the next steps: turn

$a x^{2} + b x + c ⟶ a (x + \frac{b}{2 a})^{2} + (leftover)$

so that $x$ lives in one place — the bracket $(x + \frac{b}{2 a})$ .

Step 3 — Make the bowl standard: factor out $a$

Term by term:

$a (now coefficient 1 x^{2} + half of this = \frac{b}{2 a} \frac{b}{a} x) + parked outside c$

We highlight $\frac{b}{a}$ : half of it is $\frac{b}{2 a}$ . That half is the number that will sit inside our perfect square. Hold onto it.

Step 4 — Complete the square: fold the slant into a bracket

Putting it back into the whole expression:

$a [(x + \frac{b}{2 a})^{2} - (\frac{b}{2 a})^{2}] + c = a (x + \frac{b}{2 a})^{2} the whole leftover - \frac{b ^{2}}{4 a} + c$

The bracket $(x + \frac{b}{2 a})^{2}$ is the reshaped bowl. Everything else is a constant.

Step 5 — Set $y = 0$ and isolate the square

Look hard at the right-hand side — the numerator is the star of the whole show:

$(x + \frac{b}{2 a})^{2} = \frac{b ^{2} - 4 a c call this Δ}{always > 0 4 a ^{2}}$

Step 6 — Take the square root: three futures fork here

Solving for $x$ gives the quadratic formula:

$x = centre / axis of symmetry - \frac{b}{2 a} \pm half-distance to each root \frac{Δ}{2 a}$

The two roots are a centre $\pm$ a spread. The centre is the axis of symmetry. The spread is governed entirely by $Δ$ . Now we walk each sign of $Δ$ .

Step 7 — Case $Δ > 0$ : two real roots (the branches spread apart)

Step 8 — Case $Δ = 0$ : one repeated root (branches merge at the vertex)

Step 9 — Case $Δ < 0$ : complex conjugate roots (no real crossing)

The two roots share the same centre $α = - \frac{b}{2 a}$ and differ only in the sign of the imaginary part $β = \frac{∣Δ∣}{2 a}$ — a conjugate pair. Because the coefficients $a, b, c$ are real, complex roots are forced to arrive in such pairs (the Conjugate Root Theorem).

The one-picture summary

Slide one dial and watch the vertex fall through the ground line: above → grazing → below. Each position is a sign of $Δ$ , and the crossings appear exactly as the formula predicts.

$x = \frac{- b}{2 a} \pm \frac{Δ}{2 a}, Δ = b^{2} - 4 a c, ⎩ ⎨ ⎧ Δ > 0 Δ = 0 Δ < 0 two real (crosses twice) one repeated (touches) two complex (misses)$

Recall Feynman retelling — say it like a story

We drew a U-shaped curve and asked "where does it hit the ground?" The $x$ was hiding in two spots, so we couldn't just undo it — we reshaped the bowl by completing the square until $x$ sat inside one bracket. Setting the height to zero, everything collapsed to "(bracket) $^{2}$ = a number over $4 a^{2}$ ." That number's top is $b^{2} - 4 a c$ , the discriminant. To free $x$ we square-rooted — and a square root always hands back two answers ( $\pm$ ), which is why there are two roots: a centre $- \frac{b}{2 a}$ plus-or-minus a spread $\frac{Δ}{2 a}$ . If that number is positive, the spread is real and the bowl slices the ground twice. If it's zero, the spread vanishes and both roots pile onto the vertex, which just kisses the ground. If it's negative, no real number squares to it, so the bowl floats clear of the ground and the answers become a conjugate pair of complex numbers. One number — the discriminant — tells you which of the three worlds you're in, before you ever solve.

Recall Quick self-test

Which sign of $Δ$ makes the parabola touch the x-axis exactly once? ::: $Δ = 0$ (the vertex sits on the ground; the repeated root is $- \frac{b}{2 a}$ ). Why does the quadratic formula have a $\pm$ ? ::: Square-rooting undoes a square, and both $+ s$ and $- s$ square to the same value, so both must be returned — giving two roots. For $2 x^{2} + 3 x + 5 = 0$ , what is $Δ$ and the nature of roots? ::: $Δ = - 31 < 0$ : two complex conjugate roots, $- \frac{3}{4} \pm \frac{31}{4} i$ . Why must $a \neq = 0$ throughout? ::: We divided by $a$ ; if $a = 0$ the curve is a line, not a parabola, and the discriminant idea no longer applies.

Parent: Discriminant — nature of roots (real - equal - complex) · see also Vieta's Formulas, Graphing Quadratic Functions.