Discriminant — nature of roots (real - equal - complex): Worked examples

You already met the discriminant $Δ = b^{2} - 4 a c$ in the parent note. There you saw the three headline cases. But real problems throw curveballs: a leading coefficient that might be zero, a "find $k$ " twist, a word problem, a discriminant that is a perfect square, and so on.

This page is a scenario matrix: every kind of situation this one little formula can be asked about, each with a fully worked example. Guess the answer before you read the steps — that "Forecast" habit is how you build instinct.

Definition Prerequisite reminders (so this page stands alone)

You will meet a few named tools below. Here is the one-line version of each so you never have to leave the page:

Quadratic Formula ::: the machine $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ that spits out both roots of $a x^{2} + b x + c = 0$ .
Complex Numbers ::: numbers of the form $α + β i$ where $i = - 1$ is a made-up number whose square is $- 1$ ; they let us take square roots of negatives.
Vieta's Formulas ::: two shortcuts — for $a x^{2} + b x + c = 0$ the sum of roots is $- b / a$ and the product is $c / a$ — handy for checking answers without re-solving.
Factoring Quadratics ::: writing $a x^{2} + b x + c$ as $a (x - r_{1}) (x - r_{2})$ ; only clean when roots are rational.
Conjugate Root Theorem ::: if the coefficients $a, b, c$ are ordinary real numbers, then complex roots always arrive in mirror-image pairs $α + β i$ and $α - β i$ .

The scenario matrix

Every question this topic can ask lives in one of these cells. The right column tells you which worked example nails it.

#	Scenario class	What makes it tricky	Example
A	$Δ > 0$ , perfect square	roots are rational — factors cleanly	Ex A
B	$Δ > 0$ , not a perfect square	roots are irrational (surds)	Ex B
C	$Δ = 0$	one repeated root, parabola touches axis	Ex C
D	$Δ < 0$	complex conjugate roots	Ex D
E	Find the parameter for a chosen nature	set $Δ = 0$ / $> 0$ / $< 0$ and solve	Ex E
F	Degenerate: leading coeff may be $0$	must guard $a \neq = 0$ ; the $a = 0$ branch itself	Ex F
G	Word problem (real-world)	translate first, then test $Δ$	Ex G
H	Exam twist: "always real for all values"	discriminant as an inequality in a variable	Ex H

How to walk the flowchart

The diagram below is the decision procedure every worked example secretly follows. Read it top to bottom: first ask whether it is even a quadratic, then compute $Δ$ , then branch on its sign, and finally — only on the positive branch — ask the perfect-square question that separates rational from irrational roots. The left-hand a = 0 branch (marked LIN) is not a dead end — Example F walks it concretely. Note: the node labels here are plain descriptions and do not correspond to the A–H cell letters in the table above; the letters in the table are scenario names, the boxes below are just steps.

Example A — positive & perfect square

Worked example Cell A:

$Δ > 0$ and a perfect square Problem: Describe the roots of $3 x^{2} - 7 x + 2 = 0$ .

Forecast: $Δ$ is positive here — but is it a perfect square? Guess whether the roots turn out to be nice fractions or ugly surds.

Identify $a = 3, b = - 7, c = 2$ . Why this step? The whole formula is useless until we read off signed coefficients — note $b = - 7$ , negative.
Compute $Δ = (- 7)^{2} - 4 (3) (2) = 49 - 24 = 25$ . Why this step? The sign of $Δ$ decides real vs complex; its being a square decides rational vs irrational.
Recognise $25 = 5^{2}$ , a perfect square, and $25 > 0$ . Why this step? A positive perfect-square discriminant guarantees two distinct rational roots, so Factoring Quadratics (writing it as $a (x - r_{1}) (x - r_{2})$ ) will work with no surds.
Find them: $x = \frac{7 \pm 25}{2 \cdot 3} = \frac{7 \pm 5}{6}$ , giving $x_{1} = 2, x_{2} = \frac{1}{3}$ . Why this step? $- b = - (- 7) = + 7$ — the classic sign trap flagged above.

In the figure the red curve cuts cleanly through two labelled grid points — the visual signature of rational roots.

Figure A: $y = 3 x^{2} - 7 x + 2$ (red) crosses the black x-axis at the two rational points $x = \frac{1}{3}$ and $x = 2$ — a positive, perfect-square $Δ = 25$ .

Verify: $3 (2)^{2} - 7 (2) + 2 = 12 - 14 + 2 = 0$ ✓ and $3 (\frac{1}{3})^{2} - 7 (\frac{1}{3}) + 2 = \frac{1}{3} - \frac{7}{3} + 2 = 0$ ✓. Both rational, as predicted.

Example B — positive but irrational

Worked example Cell B:

$Δ > 0$ , not a perfect square Problem: Describe the roots of $x^{2} - 4 x + 1 = 0$ .

Forecast: Positive discriminant again — but will $Δ$ come out clean? Guess before computing.

Identify $a = 1, b = - 4, c = 1$ . Why this step? Same first move always: read the signed coefficients.
Compute $Δ = (- 4)^{2} - 4 (1) (1) = 16 - 4 = 12$ . Why this step? $12 > 0$ → two real roots.
Check square-ness: $12$ is not a perfect square ( $3^{2} = 9, 4^{2} = 16$ ). Why this step? This is the difference between Cell A and Cell B — here the roots are irrational surds (see the surd definition above), so factoring over integers is hopeless; use the Quadratic Formula $x = \frac{- b \pm Δ}{2 a}$ .
Simplify the surd: $12 = 23$ , so $x = \frac{4 \pm 2 3}{2} = 2 \pm 3$ . Why this step? Pulling $12 = 4 \cdot 3 = 23$ lets the $2$ cancel the denominator cleanly.

The red curve still crosses twice, but now at irrational spots $2 \pm 3 \approx 0.27$ and $3.73$ — no tidy grid point.

Figure B: $y = x^{2} - 4 x + 1$ (red) crosses the x-axis at the two irrational points $2 - 3$ and $2 + 3$ — positive but non-square $Δ = 12$ .

Verify: Using Vieta's Formulas — sum of roots $= (2 + 3) + (2 - 3) = 4$ , which must equal $- b / a = 4$ ✓. Product $= (2)^{2} - (3)^{2} = 4 - 3 = 1 = c / a$ ✓.

Example C — zero discriminant (a figure)

Worked example Cell C:

$Δ = 0$ , repeated root Problem: Show that $4 x^{2} - 12 x + 9 = 0$ has a repeated root, and locate it. (Throughout we write $y = 4 x^{2} - 12 x + 9$ for the height of the curve above each $x$ ; a root is where $y = 0$ .)

Forecast: A repeated root means the parabola just kisses the x-axis. Guess the touch point before solving.

Identify $a = 4, b = - 12, c = 9$ . Why this step? Every discriminant problem starts by reading off the signed coefficients — here $b = - 12$ is negative, which matters when we square it.
Compute $Δ = (- 12)^{2} - 4 (4) (9) = 144 - 144 = 0$ . Why this step? $Δ = 0$ is the razor's edge — the $\pm Δ$ collapses to $\pm 0$ .
Single root: $x = \frac{- b}{2 a} = \frac{12}{8} = \frac{3}{2}$ . Why this step? With $0 = 0$ the two formula branches merge into one value, the axis of symmetry itself.
Confirm the perfect square: $4 x^{2} - 12 x + 9 = (2 x - 3)^{2}$ . Why this step? $Δ = 0$ ⇔ the quadratic is a perfect square, root of multiplicity $2$ .

Look at the red curve — it descends, grazes the axis at $x = \frac{3}{2}$ , and climbs back without crossing.

Figure C: the red parabola $y = 4 x^{2} - 12 x + 9$ dips down to touch the black x-axis at the single point $x = \frac{3}{2}$ , never crossing it — the geometric signature of $Δ = 0$ .

Verify: $(2 \cdot \frac{3}{2} - 3)^{2} = (0)^{2} = 0$ ✓. The curve's height at $x = \frac{3}{2}$ is $y = 4 (\frac{3}{2})^{2} - 12 (\frac{3}{2}) + 9 = 9 - 18 + 9 = 0$ ✓ — the lowest point (vertex) sits on the axis, so $y = 0$ there.

Example D — negative discriminant, complex roots (a figure)

Worked example Cell D:

$Δ < 0$ , complex conjugates Problem: Solve $x^{2} + 2 x + 5 = 0$ .

Forecast: Will the parabola cross the x-axis at all? If not, where do the roots hide? Guess the real part of each root.

Identify $a = 1, b = 2, c = 5$ .
Compute $Δ = 2^{2} - 4 (1) (5) = 4 - 20 = - 16$ . Why this step? $Δ < 0$ → no real roots (parabola stays above the axis), but two complex roots exist — recall " $Δ < 0$ is not 'no solution'" from the facts above.
Bring in $i$ : $- 16 = 4 i$ , where $i = - 1$ from Complex Numbers (the invented number whose square is $- 1$ ). Why this step? Real square roots of negatives don't exist, so we define $i$ to name them — that's the whole point of complex numbers.
Apply formula: $x = \frac{- 2 \pm 4 i}{2} = - 1 \pm 2 i$ . Why do the two roots mirror each other? The Conjugate Root Theorem says: because $a, b, c$ are real, the only thing that differs between the two roots is the $\pm$ in front of $Δ$ . When $Δ < 0$ that square root carries the $i$ , so one root has $+ 2 i$ and the other $- 2 i$ while their real part $- 1$ stays identical — an automatic conjugate pair. The figure shows this mirror symmetry across the real axis.

Figure D: left — the red parabola $y = x^{2} + 2 x + 5$ floats entirely above the x-axis (no real root); right — the two roots plotted as red dots in the complex plane at $- 1 \pm 2 i$ , mirror images across the horizontal (real) axis, which is exactly what "conjugate pair" means.

Verify: Plug $x = - 1 + 2 i$ : $(- 1 + 2 i)^{2} + 2 (- 1 + 2 i) + 5$ . Now $(- 1 + 2 i)^{2} = 1 - 4 i + 4 i^{2} = 1 - 4 i - 4 = - 3 - 4 i$ . Then $- 3 - 4 i - 2 + 4 i + 5 = 0$ ✓.

Example E — find the parameter for a chosen nature

Cell E is really three related tasks — one for each nature. Here we solve all three on the same family so you see the pattern: choose the sign of $Δ$ you want, then solve for the parameter.

Worked example Cell E(i): solve for

$k$ so roots are EQUAL ( $Δ = 0$ ) Problem: Find all $k$ making $x^{2} + k x + 16 = 0$ have equal roots.

Forecast: "Equal roots" is a single knife-edge condition. Guess how many values of $k$ work.

Identify $a = 1, b = k, c = 16$ . Why this step? The parameter $k$ is the coefficient $b$ , so it will land inside $Δ$ .
Equal-roots condition is $Δ = 0$ . Why this step? Only $Δ = 0$ collapses two roots into one (Cell C behaviour).
Set up: $Δ = k^{2} - 4 (1) (16) = k^{2} - 64 = 0$ . Why this step? Now the unknown lives inside $Δ$ , so the discriminant equation becomes an equation in $k$ .
Solve: $k^{2} = 64 \Rightarrow k = \pm 8$ . Two values. Why this step? $k^{2} = 64$ has two square roots — don't drop the negative.

The figure below plots $Δ$ against $k$ : it is a parabola crossing zero at exactly the two special values.

Figure E: $Δ (k) = k^{2} - 64$ (red) versus $k$ . It sits below zero for $- 8 < k < 8$ (complex roots), touches zero at $k = \pm 8$ (equal roots), and rises above zero outside (distinct real roots) — one picture covering parts (i), (ii), (iii).

Verify: For $k = 8$ : $x^{2} + 8 x + 16 = (x + 4)^{2}$ , root $- 4$ (double) ✓. For $k = - 8$ : $x^{2} - 8 x + 16 = (x - 4)^{2}$ , root $4$ (double) ✓.

Example F — the degenerate case ( $a$ might be zero)

Worked example Cell F: guard against

$a = 0$ , and the linear branch itself Problem: Discuss the roots of $(p - 2) x^{2} + 4 x + 3 = 0$ for all real $p$ — including what happens on the $a = 0$ branch of the flowchart.

Forecast: There are two regimes here: a quadratic one and a special linear one. Guess what single value of $p$ flips between them.

Spot the leading coefficient $a = p - 2$ . Why this step? The flowchart's very first question is "is $a = 0$ ?" — everything downstream depends on the answer.
The $a = 0$ (LIN) branch: if $p = 2$ , the $x^{2}$ term vanishes and the equation becomes the linear $4 x + 3 = 0$ , giving the single root $x = - \frac{3}{4}$ . There is no $Δ$ , no $\pm$ , no second root. Why this step? This anchors the flowchart's left branch: a genuinely degenerate case with exactly one root — not two, not zero. (Had the equation instead reduced to $0 \cdot x + 5 = 0$ it would have no root, and to $0 \cdot x + 0 = 0$ infinitely many — the other two degenerate outcomes to be aware of.)
The quadratic branch ( $p \neq = 2$ ): now $Δ$ is meaningful. For two distinct real roots we need $Δ > 0$ . Why this step? "Two distinct" means strictly positive, not $\geq$ .
Compute: $Δ = 4^{2} - 4 (p - 2) (3) = 16 - 12 (p - 2) = 16 - 12 p + 24 = 40 - 12 p$ . Why this step? Expand carefully — $c = 3$ , $a = p - 2$ .
Solve the inequality: $40 - 12 p > 0 \Rightarrow p < \frac{40}{12} = \frac{10}{3}$ , and keep $p \neq = 2$ . Why this step? Dividing by positive $12$ keeps the direction; the guard from step 2 still excludes $p = 2$ .

Answer: Distinct real roots for $p \in (- \infty, 2) \cup (2, \frac{10}{3})$ ; a single (linear) root at $p = 2$ ; equal roots at $p = \frac{10}{3}$ ; complex roots for $p > \frac{10}{3}$ .

Verify: $p = 2$ : equation $4 x + 3 = 0$ , one root $x = - \frac{3}{4}$ ✓. $p = 0$ : $- 2 x^{2} + 4 x + 3$ , $Δ = 16 - 4 (- 2) (3) = 40 > 0$ ✓ (two real). $p = \frac{10}{3}$ : $Δ = 40 - 12 \cdot \frac{10}{3} = 40 - 40 = 0$ ✓ (equal). $p = 4$ : $Δ = 40 - 48 = - 8 < 0$ ✓ (complex).

Example G — a real-world word problem (a figure)

Worked example Cell G: projectile height

Problem: A ball's height in metres is $h (t) = - 5 t^{2} + 20 t + 2$ ( $t$ in seconds). Does the ball ever reach a height of exactly $22$ m?

Forecast: The peak of a downward parabola is limited. Guess whether $22$ m is reachable before computing.

Translate to a quadratic: set $h (t) = 22$ , i.e. $- 5 t^{2} + 20 t + 2 = 22$ , so $- 5 t^{2} + 20 t - 20 = 0$ . Why this step? "Reaches height $22$ " means a time $t$ solving this equation exists — a real-root question.
Tidy up: divide by $- 5$ : $t^{2} - 4 t + 4 = 0$ . Why this step? Smaller integer coefficients make $Δ$ easier and can't change whether roots exist.
Compute $Δ = (- 4)^{2} - 4 (1) (4) = 16 - 16 = 0$ . Why this step? $Δ = 0$ means it reaches $22$ m at exactly one instant — the ball just kisses that height at its peak.
Find the time: $t = \frac{4}{2} = 2$ s. Why this step? $Δ = 0$ gives the single time $t = - b /2 a$ .

The red parabola touches the dashed line $h = 22$ at exactly one point — the vertex.

Figure G: the red height curve $h (t) = - 5 t^{2} + 20 t + 2$ rises to a peak that exactly grazes the dashed line $h = 22$ at $t = 2$ s, then falls — so $22$ m is reached at one instant only.

Verify: $h (2) = - 5 (4) + 20 (2) + 2 = - 20 + 40 + 2 = 22$ ✓ — units: metres, at $t = 2$ s. Reaching a higher target, say $23$ m, would give $- 5 t^{2} + 20 t - 21 = 0$ , $Δ = 400 - 420 = - 20 < 0$ → never reached, matching that $22$ m is the peak.

Example H — exam twist: "real for ALL values"

Worked example Cell H: discriminant of a discriminant

Problem: Show that $x^{2} - 2 m x + (m^{2} + 1) = 0$ has no real roots for every real number $m$ .

Forecast: The claim is that whatever $m$ you pick, roots stay complex. Guess what fixed value $Δ$ collapses to.

Identify $a = 1, b = - 2 m, c = m^{2} + 1$ . Why this step? Coefficients now contain the parameter $m$ — the discriminant will too.
Compute $Δ = (- 2 m)^{2} - 4 (1) (m^{2} + 1) = 4 m^{2} - 4 m^{2} - 4 = - 4$ . Why this step? Watch the two $4 m^{2}$ terms cancel — the $m$ -dependence disappears entirely.
Interpret: $Δ = - 4 < 0$ for every $m$ . Why this step? A discriminant that is a negative constant means the nature never changes — always two complex roots.
Conclude (by Completing the Square as a cross-check): $x^{2} - 2 m x + m^{2} + 1 = (x - m)^{2} + 1 \geq 1 > 0$ , so the left side is never zero for real $x$ . Why this step? The completed-square form visibly bottoms out at $1$ , confirming no real roots geometrically.

Verify: For $m = 3$ : $x^{2} - 6 x + 10$ , $Δ = 36 - 40 = - 4$ ✓. For $m = 0$ : $x^{2} + 1$ , $Δ = - 4$ ✓. Constant, negative, always — proven.

Recall

Recall Which cell is

$Δ > 0$ but not a perfect square? Cell B ::: two distinct irrational (surd) real roots.

Recall What extra condition must you always check in "

$a x^{2} + \dots$ " when $a$ contains a parameter? That $a \neq = 0$ ::: otherwise the equation is linear (Cell F) — it may have one root, none, or infinitely many, but the discriminant idea doesn't apply.

Recall A parameter appears inside

$Δ$ and $Δ$ simplifies to a negative constant. What does that prove? The roots are complex for every value of the parameter ::: (Cell H).

Recall Why must

$- 16$ produce a conjugate pair of roots? Because $a, b, c$ are real, the only difference between the two roots is the $\pm$ on $Δ$ ::: so one gets $+ 4 i$ , the other $- 4 i$ , same real part (Conjugate Root Theorem).