Quadratic equations — factoring, completing the square: Worked examples

Before anything, some plain-word reminders (every symbol earned):

The scenario matrix

Every quadratic you'll ever meet falls into one of these cells. The column "# roots" counts how many times the curve touches the ground.

Cell	Situation	Sign pattern / trigger	# roots	Best tool	Example
A	Monic ( $a = 1$ ), two positive roots	$b < 0$ , $c > 0$	2	factoring	Ex 1
A′	Monic, two negative roots	$b > 0$ , $c > 0$	2	factoring	Ex 2
B	Monic, roots of mixed sign	$c < 0$	2	factoring	Ex 3
C	Non-monic ( $a \neq = 1$ ), nice roots	$a c$ factors into two integers that add to $b$	2	AC method	Ex 4
D	Won't factor over integers	$Δ = b^{2} - 4 a c$ is not a perfect square	2 (irrational)	completing the square	Ex 5
E	Perfect square / repeated root	$Δ = 0$ , curve touches ground once	1	either (see note)	Ex 6
F	No real roots	$Δ < 0$ , curve floats off the ground	0	completing the square	Ex 7
G	Degenerate: $c = 0$	no constant term	2	factor out $x$	Ex 8
H	Degenerate: $b = 0$	no middle term	2 (or 0)	square-root directly	Ex 9
I	Real-world word problem	projectile / area	depends	model then solve	Ex 10
J	Exam twist: negative leading $a$	$a < 0$ , curve opens down	depends	flip sign first	Ex 11

Figure — Quadratic equations — factoring, completing the square

Look at the picture: each coloured curve is one row-class of the matrix, and each is labelled with its cells and colour. The lavender curve (cells A, A′, B, C, D, G, J) crosses the grey ground line twice → two roots. The mint curve (cell E) just touches once → one repeated root. The coral curve (cell F) never touches → no real roots. This is the single mental image behind everything below.

Cell A — monic, two positive roots

Worked example Example 1 · Solve

$x^{2} - 7 x + 12 = 0$ Forecast: $c = 12 > 0$ and $b = - 7 < 0$ . Two numbers that multiply to a positive and add to a negative must both be negative... but the roots will be the opposites of those. Guess: two positive roots. Which?

Step 1. Identify $b = - 7$ , $c = 12$ . This is monic ( $a = 1$ ). Why this step? The monic recipe (parent note, Case 1) hunts the two seeds $p, q$ that multiply to $c$ and add to $b$ . We must know them first.

Step 2. List factor pairs of $12$ : $(1, 12), (2, 6), (3, 4)$ and their negatives. Why this step? One pair will match both conditions; we test candidates.

Step 3. We need a sum of $- 7$ . Try $(- 3) + (- 4) = - 7$ ✓ and $(- 3) (- 4) = 12$ ✓. Why this step? Both conditions hold, so these are our seeds $p = - 3, q = - 4$ .

Step 4. Write $(x - 3) (x - 4) = 0$ . Why this step? $(x + p) (x + q)$ with $p = - 3, q = - 4$ rebuilds the quadratic.

Step 5. Zero Product Property: $x - 3 = 0$ or $x - 4 = 0$ , so $x = 3$ or $x = 4$ . Why this step? If a product is zero, one factor is zero — this turns one quadratic into two easy linear equations.

Verify: $3^{2} - 7 (3) + 12 = 9 - 21 + 12 = 0$ ✓ and $4^{2} - 7 (4) + 12 = 16 - 28 + 12 = 0$ ✓. Both positive, as forecast. Vieta's check: sum $3 + 4 = 7 = - b$ ✓, product $3 \cdot 4 = 12 = c$ ✓.

Cell A′ — monic, two negative roots

Worked example Example 2 · Solve

$x^{2} + 7 x + 12 = 0$ Forecast: $c = 12 > 0$ (same-sign roots) but now $b = + 7 > 0$ . Same-sign roots that add to a positive must both be positive... yet the roots are the opposites of the seeds, so guess: two negative roots. (This is Example 1's mirror image.)

Step 1. $b = 7$ , $c = 12$ (monic). Why this step? Same seed-hunt: seeds $p, q$ with $p + q = 7$ , $pq = 12$ .

Step 2. Factor pairs of $12$ : $(1, 12), (2, 6), (3, 4)$ . Why this step? Both seeds positive (product positive, sum positive), so we only need positive pairs.

Step 3. Sum must be $7$ : $3 + 4 = 7$ ✓, $3 \cdot 4 = 12$ ✓. Seeds $p = 3, q = 4$ . Why this step? Match both conditions.

Step 4. $(x + 3) (x + 4) = 0$ . Why this step? $(x + p) (x + q)$ with the positive seeds rebuilds the quadratic.

Step 5. $x + 3 = 0$ or $x + 4 = 0$ , so $x = - 3$ or $x = - 4$ . Why this step? Zero Product Property — positive seeds give negative roots, as forecast.

Verify: $(- 3)^{2} + 7 (- 3) + 12 = 9 - 21 + 12 = 0$ ✓; $(- 4)^{2} + 7 (- 4) + 12 = 16 - 28 + 12 = 0$ ✓. Vieta: sum $- 3 - 4 = - 7 = - b$ ✓, product $(- 3) (- 4) = 12 = c$ ✓.

Cell B — mixed-sign roots

Worked example Example 3 · Solve

$x^{2} + 2 x - 15 = 0$ Forecast: $c = - 15 < 0$ . A negative product means the two numbers have opposite signs — so one root positive, one negative. Which is bigger in size? Since they must add to $- b = - 2$ (a negative), the negative root must outweigh the positive one — so the negative root has the larger absolute value.

Step 1. $b = 2$ , $c = - 15$ (monic again). Why this step? Same seed-hunt; record the targets $p + q = 2$ , $pq = - 15$ .

Step 2. Factor pairs of $- 15$ (opposite signs): $(1, - 15), (3, - 5), (5, - 3), (15, - 1)$ . Why this step? Only opposite-sign pairs can multiply to a negative.

Step 3. Sum must equal $2$ : $5 + (- 3) = 2$ ✓, $5 \cdot (- 3) = - 15$ ✓. Seeds $p = 5, q = - 3$ . Why this step? Match both conditions.

Step 4. $(x + 5) (x - 3) = 0$ . Why this step? $(x + p) (x + q)$ with $p = 5, q = - 3$ rebuild the quadratic.

Step 5. $x = - 5$ or $x = 3$ . Why this step? Zero Product Property — note the signs flip from inside the brackets.

Verify: $(- 5)^{2} + 2 (- 5) - 15 = 25 - 10 - 15 = 0$ ✓; $3^{2} + 2 (3) - 15 = 9 + 6 - 15 = 0$ ✓. One negative, one positive, and $∣ - 5∣ > ∣3∣$ — the negative root is indeed larger in size, exactly as forecast.

Cell C — non-monic, AC method

Worked example Example 4 · Solve

$3 x^{2} - 10 x + 8 = 0$ Forecast: $a = 3$ so it's non-monic — factoring needs the AC trick. Product $a c = 24$ , sum $- 10$ : both negative. Guess two rational roots.

Step 1. Compute $a c = 3 \cdot 8 = 24$ . Why this step? AC means "multiply $a$ times $c$ ." This folds the leading coefficient into one target number so we can hunt a multiply-and-add pair like a monic (see the AC intuition box above).

Step 2. Two numbers multiplying to $24$ , adding to $- 10$ : $(- 4) (- 6) = 24$ , $- 4 - 6 = - 10$ ✓. Why this step? These are the amounts that will split the middle term into two groupable strips.

Step 3. Rewrite: $3 x^{2} - 4 x - 6 x + 8 = 0$ . Why this step? Splitting $- 10 x$ into $- 4 x - 6 x$ prepares grouping — the sum is unchanged so the equation is identical.

Step 4. Group: $(3 x^{2} - 4 x) + (- 6 x + 8) = 0 \Rightarrow x (3 x - 4) - 2 (3 x - 4) = 0$ . Why this step? Pulling a common factor out of each pair exposes the shared binomial $(3 x - 4)$ — the "shared side" the two strips glue along.

Step 5. $(3 x - 4) (x - 2) = 0$ . Why this step? The common binomial factors out — the hidden factorization appears.

Step 6. $3 x - 4 = 0 \Rightarrow x = \frac{4}{3}$ ; $x - 2 = 0 \Rightarrow x = 2$ . Why this step? Zero Product Property on each factor.

Verify: $3 (\frac{4}{3})^{2} - 10 (\frac{4}{3}) + 8 = 3 \cdot \frac{16}{9} - \frac{40}{3} + 8 = \frac{16}{3} - \frac{40}{3} + \frac{24}{3} = 0$ ✓; $3 (4) - 20 + 8 = 0$ ✓.

Cell D — won't factor: complete the square

Worked example Example 5 · Solve

$x^{2} - 6 x + 2 = 0$ Forecast: The discriminant is $Δ = (- 6)^{2} - 4 (1) (2) = 36 - 8 = 28$ , which is not a perfect square. So integer factoring fails and we expect irrational roots near $x = 3$ .

Step 1. Move the constant across: $x^{2} - 6 x = - 2$ . Why this step? Completing the square needs the $x^{2}$ and $x$ terms alone so we can build a perfect square around them.

Step 2. Take half of $b = - 6$ , then square: $(\frac{- 6}{2})^{2} = 9$ . Why this step? A perfect square $(x + k)^{2}$ has middle term $2 k$ ; here $2 k = - 6$ so $k = - 3$ and the missing corner is $k^{2} = 9$ .

Step 3. Add $9$ to both sides: $x^{2} - 6 x + 9 = - 2 + 9 = 7$ . Why this step? Balance — whatever we add left, we add right, or the equation breaks.

Step 4. Fold the left into a square: $(x - 3)^{2} = 7$ . Why this step? $x^{2} - 6 x + 9$ is exactly $(x - 3)^{2}$ — that was the whole point.

Step 5. Square-root both sides: $x - 3 = \pm 7$ . Why this step? Undoing a square gives two signs, capturing both roots at once. (Note $7$ matches $Δ /2 = 28 /2 = 7$ .)

Step 6. $x = 3 \pm 7$ . Why this step? Add $3$ to isolate $x$ .

Verify: Roots are $3 + 7 \approx 5.646$ and $3 - 7 \approx 0.354$ , both near $3$ as forecast. Vieta's check: sum $= (3 + 7) + (3 - 7) = 6 = - b$ ✓, product $= (3)^{2} - (7)^{2} = 9 - 7 = 2 = c$ ✓.

Cell E — repeated root (curve just kisses the ground)

Cell F — no real roots (curve never lands)

Worked example Example 7 · Solve

$x^{2} + 2 x + 5 = 0$ Forecast: $Δ = b^{2} - 4 a c = 2^{2} - 4 (1) (5) = 4 - 20 = - 16 < 0$ . A negative discriminant means no real answers — the curve floats entirely above the ground.

Step 1. Move the constant across: $x^{2} + 2 x = - 5$ . Why this step? Isolate the $x^{2}$ and $x$ terms so we can build a square around them.

Step 2. Half of $b = 2$ is $1$ ; its square is $1$ . Add and subtract that $1$ on the left: $x^{2} + 2 x + 1 - 1 = - 5.$ Why this step? Writing the $+ 1 - 1$ explicitly shows we changed nothing — we only reorganized, so the perfect square can emerge.

Step 3. The first three terms are a perfect square: $x^{2} + 2 x + 1 = (x + 1)^{2}$ . So: $(x + 1)^{2} - 1 = - 5.$ Why this step? $(x + 1)^{2}$ expands to $x^{2} + 2 x + 1$ — exactly the grouped part.

Step 4. Add the leftover $1$ to the right: $(x + 1)^{2} = - 4$ . Why this step? Isolate the square to inspect it.

Step 5. A real number squared is never negative, so no real $x$ works. Why this step? $(x + 1)^{2} \geq 0$ always, but the right side is $- 4$ — impossible for real $x$ .

Verify: The vertex (turning point) sits at $x = - 1$ , where $y = (- 1)^{2} + 2 (- 1) + 5 = 4 > 0$ . Lowest point is above ground ⇒ zero real roots, matching the coral curve that never dips below the line.

Cell G — degenerate: $c = 0$

Cell H — degenerate: $b = 0$

Cell I — real-world word problem

Worked example Example 10 · A ball is thrown upward; its height in metres after

$t$ seconds is $h (t) = - 5 t^{2} + 20 t$ . When does it hit the ground? Forecast: "Hits the ground" means $h = 0$ . It leaves the ground at $t = 0$ and lands later — expect two times, one being $t = 0$ .

Step 1. Set $h = 0$ : $- 5 t^{2} + 20 t = 0$ . Why this step? Ground level is height zero; solving gives the launch and landing times.

Step 2. Factor out $- 5 t$ : $- 5 t (t - 4) = 0$ . Why this step? No constant term (cell G pattern) — factoring $t$ is cleanest.

Step 3. $t = 0$ (launch) or $t - 4 = 0 \Rightarrow t = 4$ s (landing). Why this step? Zero Product gives both physical times; $- 5$ can't be zero.

Step 4. We want the landing time, so $t = 4$ seconds. Why this step? $t = 0$ is the throw itself; the ball returns to the ground at $t = 4$ .

Verify: $h (4) = - 5 (16) + 20 (4) = - 80 + 80 = 0$ m ✓. Units: seconds in, metres out — consistent. The vertex (peak) is halfway between the roots, at $t = 2$ s, height $h (2) = - 20 + 40 = 20$ m — a sensible top.

Reading the figure: the lavender curve is the ball's height $h (t)$ . The two coral dots on the ground line are the roots we just found — launch at $t = 0$ and landing at $t = 4$ s. The butter dot at the top is the vertex $(t = 2, h = 20)$ , sitting exactly halfway between the two roots. Notice the whole shape is symmetric about $t = 2$ : that vertical mirror line through the vertex is why the peak time is the average of the two ground times, a fact you'll formalize in 2.3.4-Parabolas-and-vertex-form.

Cell J — exam twist: negative leading coefficient

Worked example Example 11 · Solve

$- x^{2} + 4 x + 5 = 0$ Forecast: $a = - 1 < 0$ , the parabola opens downward. Multiply through by $- 1$ first to make life easy; then factor. Expect mixed-sign roots.

Step 1. Multiply every term by $- 1$ : $x^{2} - 4 x - 5 = 0$ . Why this step? A leading $- 1$ hides factor patterns; flipping all signs keeps the same roots (multiplying an equation by a nonzero number never changes its solutions).

Step 2. Monic factoring: need seeds $p, q$ with product $- 5$ , sum $- 4$ : $(- 5) (+ 1) = - 5$ , $- 5 + 1 = - 4$ ✓. Why this step? Standard Cell-A/B recipe.

Step 3. $(x - 5) (x + 1) = 0$ . Why this step? $p = - 5, q = 1$ rebuild the flipped quadratic.

Step 4. $x = 5$ or $x = - 1$ . Why this step? Zero Product Property.

Verify: In the original equation: $- (5)^{2} + 4 (5) + 5 = - 25 + 20 + 5 = 0$ ✓; $- (- 1)^{2} + 4 (- 1) + 5 = - 1 - 4 + 5 = 0$ ✓. Flipping signs preserved the roots exactly, as claimed.

Recall Quick self-test

Which cell does $x^{2} + 9 = 0$ land in, and how many real roots? ::: Cell H sub-case ( $b = 0$ , gives $x^{2} = - 9$ ) → no real roots. Which tool fits $x^{2} - 6 x + 2 = 0$ and why? ::: Its discriminant $Δ = 28$ isn't a perfect square, so integer factoring fails (no whole-number pair multiplies to $2$ and adds to $- 6$ ). The quadratic formula would work too, but completing the square is the cleanest here — it derives those irrational roots $3 \pm 7$ directly and shows the geometry (vertex at $x = 3$ ). A curve that touches the ground exactly once is which cell? ::: Cell E — a repeated (double) root, $Δ = 0$ , a perfect-square trinomial. For $- 5 t^{2} + 20 t = 0$ , why is one answer discarded? ::: $t = 0$ is the launch instant; the physical "landing" is $t = 4$ s. In a mixed-sign quadratic, which root is larger in size? ::: The one whose sign matches $- b$ ; e.g. in Example 3 the roots sum to $- 2$ , so the negative root ( $- 5$ ) outweighs the positive one ( $3$ ).

Where to go next

Turn the pattern of Example 5/7 into a formula: 2.1.17-Quadratic-formuland-discriminant (the discriminant $Δ = b^{2} - 4 a c$ decides cells D vs E vs F).
See these roots as vertex geometry: 2.3.4-Parabolas-and-vertex-form.
Bigger factoring toolkit: 3.2.1-Polynomials-factoring-techniques.
The linear equations each factor collapses to: 2.1.15-Linear-equations-and-inequalities.
Two curves at once: 4.1.2-Solving-systems-graphically.
Back to the parent: 2.1.16 Quadratic equations — factoring, completing the square (Hinglish).