Worked examples — Quadratic formula — derivation by completing the square

This is the worked-example deep dive for Quadratic formula — derivation by completing the square. The parent note built the formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ from scratch. Here we don't re-derive it — we stress-test it against every kind of quadratic you can meet, so no exam problem is a surprise.

Before anything else, one reminder of what each letter means, because we will match them fast:

And the one number that decides the shape of the answer:

The scenario matrix

Every quadratic you will ever be handed falls into exactly one of these cells. The last column names the example that covers it.

Cell	What makes it special	Where the trap is	Example
A — clean split	$a=1$, $\Delta >0$ perfect square	none, warm-up	Ex 1
B — negative $b$	$b<0$ so $-b$ flips sign	the $-b$ sign flip	Ex 2
C — negative $c$	$c<0$ so $-4ac$ becomes $+$	the $-4ac$ double negative	Ex 3
D — repeated root	$\Delta =0$	$\pm 0$ gives one answer	Ex 4
E — irrational roots	$\Delta >0$ but not a perfect square	leave ​ in surd form	Ex 5
F — complex roots	$\Delta <0$	$\sqrt{-k}=i\sqrt{k}$	Ex 6
G — leading $a\ne 1$, negative	$a<0$	dividing by negative $2a$	Ex 7
H — word problem (geometry)	pull $a,b,c$ from a sentence	reject the impossible root	Ex 8
I — exam twist (parameter)	an unknown $k$ inside	set $\Delta =0$ and solve for $k$	Ex 9

We now walk cell by cell.

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Forecast: Guess — will the roots be whole numbers, or ugly? (Try: $3$ and $4$ multiply to $12$ and add to $7$… hold that thought.)

1. Match coefficients. $a=1$, $b=-7$, $c=12$. Why this step? Everything downstream is just plugging these three into the formula; a wrong match poisons the whole answer.
2. Compute $\Delta$. $\Delta =(-7{)}^2-4(1)(12)=49-48=1$. Why this step? $\Delta =1>0$ and it's a perfect square ($\sqrt{1}=1$), so I already know: two nice rational roots.
3. Apply the formula. $x=\frac{-(-7)\pm \sqrt{1}}{2(1)}=\frac{7\pm 1}{2}$ Why this step? $-b=-(-7)=+7$ — the first sign flip. The $\pm$ then splits into two.
4. Split. $x=\frac{7+1}{2}=4$ or $x=\frac{7-1}{2}=3$.

Verify: $4^2-7(4)+12=16-28+12=0$ ✓ and $3^2-7(3)+12=9-21+12=0$ ✓. My forecast ($3$ and $4$) was right.

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Forecast: $b$ is $-9$. Will $-b$ be $-9$ or $+9$? Decide before reading on.

1. Match. $a=1$, $b=-9$, $c=20$. Why this step? The whole point of this cell is to see the sign of $b$ explicitly. Write it as $b=-9$, not "9 with a minus lying around."
2. $\Delta$. $\Delta =(-9{)}^2-4(1)(20)=81-80=1$. Why this step? Squaring kills the sign: $(-9{)}^2=+81$. A very common slip is writing $-81$.
3. Formula. $x=\frac{-(-9)\pm \sqrt{1}}{2}=\frac{9\pm 1}{2}$ Why this step? $-b=-(-9)=+9$. This is the trap: the numerator starts positive even though $b$ was negative.
4. Split. $x=5$ or $x=4$.

Verify: $5^2-9(5)+20=25-45+20=0$ ✓; $4^2-9(4)+20=16-36+20=0$ ✓.

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Forecast: Since $c=-2$ is negative, will $-4ac$ add to or subtract from $b^2$?

1. Match. $a=3$, $b=5$, $c=-2$. Why this step? Keep the minus glued to the $2$.
2. $\Delta$ carefully. $-4ac=-4(3)(-2)=+24$. So $\Delta =5^2+24=25+24=49.$ Why this step? Two negatives ($-4$ times a negative $c$) make a plus. A negative $c$ grows the discriminant — guaranteeing real roots.
3. Formula. $x=\frac{-5\pm \sqrt{49}}{2(3)}=\frac{-5\pm 7}{6}$ Why this step? $\sqrt{49}=7$, denominator $2a=6$.
4. Split. $x=\frac{-5+7}{6}=\frac{2}{6}=\frac{1}{3}$ or $x=\frac{-5-7}{6}=\frac{-12}{6}=-2$.

Verify: $3{\left(\frac{1}{3}\right)}^2+5\left(\frac{1}{3}\right)-2=\frac{1}{3}+\frac{5}{3}-2=2-2=0$ ✓; $3(-2{)}^2+5(-2)-2=12-10-2=0$ ✓.

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Forecast: How many distinct answers — two, one, or none?

1. Match. $a=4$, $b=-12$, $c=9$.
2. $\Delta$. $\Delta =(-12{)}^2-4(4)(9)=144-144=0$. Why this step? $\Delta =0$ is the razor's edge: the $\pm \sqrt{0}=\pm 0$ collapses the two roots into one. Geometrically the parabola just kisses the $x$-axis (see figure).
3. Formula. $x=\frac{-(-12)\pm 0}{2(4)}=\frac{12}{8}=\frac{3}{2}$ Why this step? With $\sqrt{\Delta }=0$, the $\pm$ does nothing.
4. Recognise the perfect square. $4x^2-12x+9=(2x-3{)}^2$, zero when $2x-3=0$, i.e. $x=\frac{3}{2}$. Same answer.

Verify: $4{\left(\frac{3}{2}\right)}^2-12\left(\frac{3}{2}\right)+9=9-18+9=0$ ✓.

The green parabola ($\Delta =0$) touches the axis at one point — that's why there is one repeated root. Compare it to the blue parabola crossing twice and the red one floating clear.

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Forecast: $\Delta$ won't be a perfect square this time. Do we round, or leave the root symbolic?

1. Match. $a=1$, $b=-4$, $c=1$.
2. $\Delta$. $\Delta =(-4{)}^2-4(1)(1)=16-4=12$. Why this step? $12$ is positive but not a perfect square, so the answers are irrational — keep them exact.
3. Formula and simplify the surd. $x=\frac{4\pm \sqrt{12}}{2}.$ Since $\sqrt{12}=\sqrt{4\cdot 3}=2\sqrt{3}$, $x=\frac{4\pm 2\sqrt{3}}{2}=2\pm \sqrt{3}.$ Why this step? Pull the perfect-square factor $4$ out of $12$, then cancel the common factor $2$ top and bottom. Never write a rounded decimal as the answer in exact algebra.
4. The two roots. $x=2+\sqrt{3}$ or $x=2-\sqrt{3}$.

Verify: their sum is $4=-b/a$ ✓ and product is $(2{)}^2-(\sqrt{3}{)}^2=4-3=1=c/a$ ✓ (a Vieta check).

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Forecast: $\Delta$ will be negative. Does that mean "no answers", or "no real answers"?

1. Match. $a=1$, $b=-2$, $c=5$.
2. $\Delta$. $\Delta =(-2{)}^2-4(1)(5)=4-20=-16$. Why this step? Negative $\Delta$ → the parabola never touches the axis → no real crossings, but two complex roots.
3. Introduce $i$. By definition $i=\sqrt{-1}$, so $\sqrt{-16}=\sqrt{16}\cdot \sqrt{-1}=4i$. Why this step? This is why we invented $i$ — to give $\sqrt{\text{negative}}$ a name so the formula still works.
4. Formula. $x=\frac{2\pm 4i}{2}=1\pm 2i.$

Verify: sum $=2=-b/a$ ✓; product $=(1{)}^2-(2i{)}^2=1-(-4)=5=c/a$ ✓.

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Forecast: A negative $a$ opens the parabola downward. Will that change the roots, or just flip the picture?

1. Match — signs matter most here. $a=-2$, $b=3$, $c=5$. Why this step? One tempting shortcut is multiplying the equation by $-1$ first: $2x^2-3x-5=0$. Roots are unchanged (dividing an equation by a nonzero number never moves its roots). We'll do it the direct way to practise a negative $2a$.
2. $\Delta$. $\Delta =3^2-4(-2)(5)=9+40=49$. Why this step? $-4ac=-4(-2)(5)=+40$ — again a negative coefficient boosts $\Delta$.
3. Formula with negative denominator. $x=\frac{-3\pm \sqrt{49}}{2(-2)}=\frac{-3\pm 7}{-4}.$ Why this step? $2a=-4$. Divide both branches by $-4$ — a sign trap.
4. Split. $x=\frac{-3+7}{-4}=\frac{4}{-4}=-1$ or $x=\frac{-3-7}{-4}=\frac{-10}{-4}=\frac{5}{2}$.

Verify: $-2(-1{)}^2+3(-1)+5=-2-3+5=0$ ✓; $-2{\left(\frac{5}{2}\right)}^2+3\left(\frac{5}{2}\right)+5=-\frac{25}{2}+\frac{15}{2}+5=-5+5=0$ ✓.

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Forecast: We'll get two algebraic roots but only one can be a real width. Which sign survives?

1. Name the unknown. Let width $=w$ metres. Then length $=w+3$. Why this step? Turning the sentence into one variable is the whole battle in word problems.
2. Build the equation from area. Area $=$ width $\times$ length: $w(w+3)=40\Rightarrow w^2+3w-40=0.$ Why this step? Now it's a standard quadratic; match $a=1$, $b=3$, $c=-40$.
3. $\Delta$. $\Delta =3^2-4(1)(-40)=9+160=169$. Why this step? $c<0$ again forces $-4ac$ positive; $\sqrt{169}=13$, nice.
4. Formula. $w=\frac{-3\pm 13}{2}\Rightarrow w=5\text{or}w=-8.$
5. Reject the impossible root. A width can't be $-8\text{ m}$, so $w=5\text{ m}$. Why this step? The algebra gives all mathematical roots; the physics (a length is positive) filters them.

Verify: width $5$, length $8$, area $5\times 8=40{\text{ m}}^2$ ✓ and $8=5+3$ ✓ (units: metres × metres = m², correct).

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Forecast: "Exactly one root" is a keyword. What must $\Delta$ equal?

1. Translate the wording into $\Delta$. One repeated root ⟺ $\Delta =0$. Why this step? This is the exam's disguise: instead of solving the quadratic, they hide $\Delta$ behind English.
2. Match and write $\Delta$. $a=1$, $b=k$, $c=9$, so $\Delta =k^2-4(1)(9)=k^2-36.$
3. Set $\Delta =0$ and solve for $k$. $k^2-36=0\Rightarrow k^2=36\Rightarrow k=\pm 6.$ Why this step? Both signs are valid parameters — the question asked about $k$, and $k^2$ hides the sign.
4. Report both. $k=6$ or $k=-6$.

Verify: with $k=6$: $x^2+6x+9=(x+3{)}^2$, root $x=-3$ (single) ✓. With $k=-6$: $x^2-6x+9=(x-3{)}^2$, root $x=3$ (single) ✓.

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Recall Quick self-test on the matrix

Which cell does $2x^2-4x+9=0$ fall into, and why? ::: $\Delta =16-72=-56<0$ → Cell F, complex roots. In $x^2-3x-10=0$, what is $-4ac$? ::: $-4(1)(-10)=+40$ (Cell C double-negative). Why do word-problem quadratics often discard a root? ::: A negative length/time/count is physically impossible, so only the positive root survives (Cell H). "$\Delta =0$" corresponds to which picture? ::: The parabola just touches the $x$-axis at one point (Cell D).

Connections

• Discriminant and nature of roots — the sign of $\Delta$ drives Cells D, E, F.
• Complex numbers and quadratic equations — Cell F needs $i$.
• Vieta's formulas for sum and product of roots — the sum/product checks in Ex 5 and 6.
• Perfect square trinomials — Cell D's repeated root is a perfect square.
• Parabola and its vertex form — the figures here are these parabolas.
• Factoring quadratics — an alternate route for Cells A, B, C.
• Quadratic inequalities — next step once you know where roots sit.
• Completing the square technique — the engine behind the parent derivation.