Quadratic formula — derivation by completing the square: Question bank

Reminder of the shape everything below refers to:

True or false — justify

Every item is a claim. Decide true/false and say why before revealing.

TF1. "If

a = 0

, the quadratic formula still works, we just get a simpler answer."

False — with

a = 0

you divide by

2 a = 0

, which is undefined; the equation isn't quadratic at all, it's the linear equation

b x + c = 0

TF2. "A quadratic equation always has exactly two real solutions."

False — it has two roots counting multiplicity in the complex numbers; over the reals it may have two, one (repeated), or zero, depending on the sign of

Δ

. See Discriminant and nature of roots.

TF3. "If

Δ = 0

the equation has no solution because

\pm 0

adds nothing."

False —

\pm 0 = 0

still gives a solution, namely the single repeated root

x = - \frac{b}{2 a}

; "one root" is not "no root."

TF4. "

Δ < 0

means the equation is broken and has no answers."

False — it has no real answers, but two complex conjugate roots

- \frac{b}{2 a} \pm \frac{∣Δ∣}{2 a} i

; see Complex numbers and quadratic equations.

TF5. "The

\pm

in the formula is optional — you can just pick

+

False — the

\pm

encodes the two square roots of a positive number (

y^{2} = k \Rightarrow y = \pm k

); dropping one throws away a genuine root.

TF6. "Completing the square and the quadratic formula are two different, unrelated methods."

False — the formula IS completing the square done once on the general

a x^{2} + b x + c

; every specific completing-the-square problem is just re-deriving the same formula with numbers.

TF7. "You can complete the square only after making the leading coefficient equal to 1."

True in spirit — the clean

(x + h)^{2}

pattern needs coefficient 1 on

x^{2}

, so you first divide by

a

(legal since

a \neq = 0

); skipping this makes the perfect-square number wrong.

TF8. "The number you add to complete the square,

(\frac{b}{2 a})^{2}

, must be added to both sides."

True — adding it to only the left changes the equation's value; equality is preserved only if the same amount lands on both sides.

TF9. "

4 a^{2} = 2 a

always."

False —

4 a^{2} = 2∣ a ∣

, which equals

2 a

only when

a > 0

; the formula survives anyway because the

\pm

absorbs the sign of

a

TF10. "If a quadratic has two identical roots, its graph crosses the

x

-axis twice at the same spot."

False — it touches (is tangent to) the axis at one point, the vertex; it never crosses. See Parabola and its vertex form.

Spot the error

Each line contains a flawed step. Name the mistake and give the correction.

SE1. "For

x^{2} - 5 x + 6 = 0

, I set

b = 5

- b = - 5

Error:

b

is the coefficient with its sign, so

b = - 5

and

- b = + 5

; reading the sign off wrong flips the whole numerator.

SE2. "

x = - b \pm b^{2} - 4 a c

, then I'll divide by

2 a

at the end if I remember."

Error: the entire numerator sits over

2 a

from the start, including the

\pm

term; write it as one fraction so nothing escapes the denominator.

SE3. "With

c = - 4

b^{2} - 4 a c = 49 - 4 (2) (- 4) = 49 - 32 = 17

Error:

- 4 a c = - 4 (2) (- 4) = + 32

, so it's

49 + 32 = 81

; a negative

c

flips the subtraction into an addition.

SE4. "

(x + \frac{b}{2 a})^{2} = \frac{b ^{2} - 4 a c}{4 a ^{2}}

, so

x + \frac{b}{2 a} = \frac{b ^{2} - 4 a c}{2 a}

Error: they forgot the square root — taking a square root of the left cancels the square, so the right must also be rooted, giving

\frac{b ^{2} - 4 a c}{2 a}

SE5. "

- \frac{c}{a} + \frac{b ^{2}}{4 a ^{2}} = \frac{b ^{2} - c}{4 a ^{2}}

Error: the two fractions have different denominators;

- \frac{c}{a} = - \frac{4 a c}{4 a ^{2}}

, so the sum is

\frac{b ^{2} - 4 a c}{4 a ^{2}}

, not

\frac{b ^{2} - c}{4 a ^{2}}

SE6. "To complete the square in

x^{2} + 6 x

, I add

6^{2} = 36

Error: you add

(\frac{6}{2})^{2} = 9

, not

36

; the rule is half the linear coefficient, then square. See Perfect square trinomials.

SE7. "

2 x^{2} + 7 x - 4 = 0

: I completed the square without dividing by 2 first, using

(\frac{7}{2})^{2}

Error: after dividing by

a = 2

the linear coefficient is

\frac{7}{2}

, so you need

(\frac{7}{4})^{2}

; using

\frac{7}{2}

pretends the leading coefficient was 1.

SE8. "

- 16 = - 4

Error:

- 4

is a real number whose square is

+ 16

, not

- 16

; the correct value is

4 i

where

i^{2} = - 1

Why questions

Answer the reason, not just the fact.

WHY1. Why must we divide by

a

before completing the square?

Because the perfect-square pattern

(x + h)^{2} = x^{2} + 2 h x + h^{2}

starts with a bare

x^{2}

(coefficient 1); dividing by

a

produces that leading 1 so the pattern applies.

WHY2. Why is the completing-the-square number exactly

(\frac{k}{2})^{2}

and not something else?

Matching

x^{2} + k x

x^{2} + 2 h x + h^{2}

forces

2 h = k

, so

h = \frac{k}{2}

, and the missing constant is

h^{2} = (\frac{k}{2})^{2}

WHY3. Why does the

\pm

appear only after taking the square root, never before?

Squaring loses sign information (

(+ 3)^{2} = (- 3)^{2} = 9

), so undoing the square must restore both possibilities, which is what

\pm

records.

WHY4. Why does the discriminant, sitting under a root, decide the number of real roots?

A real square root exists only for non-negative inputs, so

Δ > 0

gives two real values,

Δ = 0

gives one, and

Δ < 0

gives none in the reals.

WHY5. Why is

a \neq = 0

built into the very definition of a quadratic?

a = 0

the

x^{2}

term vanishes, leaving a linear equation; keeping

a \neq = 0

guarantees there really is a squared term to solve.

WHY6. Why does

Δ = 0

correspond to the parabola touching the axis rather than missing or crossing it?

The two roots collapse to one value

- \frac{b}{2 a}

, the vertex's

x

-coordinate, so the single contact point is exactly where the curve turns — tangent to the axis. See Parabola and its vertex form.

WHY7. Why can we write

2 a

in the denominator instead of the more honest

2∣ a ∣

The

\pm

already produces both signs of the root, so whether

2∣ a ∣

2 a

scales it, the same pair of roots emerges; the sign of

a

just relabels which is "

+

WHY8. Why does completing the square "untangle" the fact that

x

appears twice?

It fuses the

x^{2}

and

x

terms into one

(x + h)^{2}

block, so

x

appears in a single place and can be freed by one square root.

Edge cases

Boundary and degenerate inputs — the scenarios that break careless formulas.

EC1. What happens to the formula when

c = 0

(equation

a x^{2} + b x = 0

Δ = b^{2} \geq 0

so roots are always real:

\frac{- b \pm ∣ b ∣}{2 a}

, giving

x = 0

and

x = - \frac{b}{a}

; factoring

x (a x + b) = 0

confirms this. See Factoring quadratics.

EC2. What happens when

b = 0

(equation

a x^{2} + c = 0

The formula gives

x = \pm \frac{- 4 a c}{2 a} = \pm - \frac{c}{a}

; roots are real and symmetric about 0 when

- \frac{c}{a} \geq 0

, and purely imaginary otherwise.

EC3. What if both

b = 0

and

c = 0

Then

a x^{2} = 0

forces

x = 0

as a double root;

Δ = 0

, and the parabola's vertex sits on the origin.

EC4. What is the largest

Δ

can tell you if it turns out to be a perfect square (like 81)?

A perfect-square

Δ

signals rational roots, meaning the quadratic also factors nicely over the integers/rationals. See Factoring quadratics.

EC5. If

a < 0

, does the formula still hold without any adjustment?

Yes — nothing in the derivation assumed

a > 0

; the

\pm

and the algebra work for negative

a

, the parabola simply opens downward.

EC6. What does it mean if the sum of the two roots equals

- \frac{b}{a}

regardless of

Δ

's sign?

The roots (even complex ones) always sum to

- \frac{b}{a}

and multiply to

\frac{c}{a}

; the

\pm

terms cancel on adding and combine on multiplying. See Vieta's formulas for sum and product of roots.

EC7. When two roots are complex, why must they be conjugates rather than any two complex numbers?

Because

a, b, c

are real, the only complex part comes from

\pm Δ

with

Δ < 0

; the

\pm

forces

- 1 + 2 i

and

- 1 - 2 i

style pairs that mirror across the real axis. See Complex numbers and quadratic equations.

EC8. Does a repeated root (

Δ = 0

) count as "two solutions" when solving an inequality like

a x^{2} + b x + c > 0

The single root splits the line into regions, but the quadratic never changes sign there (it only touches zero), which matters for Quadratic inequalities.

Recall Quick self-audit

If you missed any "Why" item, re-derive the formula once by hand from $a x^{2} + b x + c = 0$ — every trap on this page is one careless line in that derivation.