Completing the Square

Expand \( (x + 3)^2 \; – \; 4 \): we get \( (x + 3)(x + 3) \; – \; 4 = x^2 + 6x + 9 \; – \; 4 = x^2 + 6x + 5 \), confirming the two forms are identical. The key insight is that \( (x + 3)^2 \) creates a “phantom” square of 9 that isn’t in the original equation.

This visual shows why we subtract. Forming the square \( (x+3)^2 \) physically creates an extra \( 3 \times 3 \) corner. To balance the equation, we must subtract this 9 from the original constant (+5), resulting in \( 5 – 9 = -4 \).

Complete the square: half of 8 is 4, so \( (x \; – \; 4)^2 = x^2 \; – \; 8x + 16 \). Therefore \( x^2 \; – \; 8x + 20 = (x \; – \; 4)^2 \; – \; 16 + 20 = (x \; – \; 4)^2 + 4 \). Since any squared term is greater than or equal to zero, the smallest value \( (x \; – \; 4)^2 \) can take is 0 (when \( x = 4 \)), making the minimum of the whole expression \( 0 + 4 = 4 \). This targets the “constant equals minimum” misconception — the 20 in the original form is not the minimum value; the completed square form reveals the true minimum.

Verify: when \( x = 4 \), we get \( 16 \; – \; 32 + 20 = 4 \). When \( x = 0 \), we get \( 0 \; – \; 0 + 20 = 20 \), which is larger than 4, confirming 20 is not the minimum.

Expand \( (x + 14)^2 \): this gives \( x^2 + 28x + 196 \). Adding 40 gives \( x^2 + 28x + 236 \), which has a coefficient of \( 28x \) — not the \( 14x \) we started with. The error is the “use the full coefficient” misconception: placing the entire coefficient of \( x \) inside the bracket rather than halving it. When you square \( (x + k) \), the coefficient of \( x \) in the expansion is \( 2k \), not \( k \). So to produce \( 14x \), we need \( 2k = 14 \), giving \( k = 7 \).

The correct completed square form is \( x^2 + 14x + 40 = (x + 7)^2 \; – \; 49 + 40 = (x + 7)^2 \; – \; 9 \). We can verify: \( (x + 7)^2 \; – \; 9 = x^2 + 14x + 49 \; – \; 9 = x^2 + 14x + 40 \).

The minimum value of \( (x \; – \; 3)^2 \) occurs when \( x \; – \; 3 = 0 \), i.e. when \( x = 3 \). At this point, \( y = 0 + 2 = 2 \), so the turning point is \( (3, \; 2) \). The “wrong sign for the turning point” misconception comes from reading the \( -3 \) directly from the bracket and writing \( x = -3 \). The sign must be reversed because the bracket says \( (x \; – \; 3) \), which equals zero when \( x = +3 \).

We can verify by substitution: when \( x = -3 \), \( y = (-3 \; – \; 3)^2 + 2 = 36 + 2 = 38 \), which is clearly not the minimum. When \( x = 3 \), \( y = (3 \; – \; 3)^2 + 2 = 0 + 2 = 2 \).

Example: \( x^2 + 10x + 30 = (x + 5)^2 + 5 \)

Another: \( x^2 + 10x + 25 = (x + 5)^2 \)  (i.e. \( q = 0 \))

Creative: \( x^2 + 10x = (x + 5)^2 \; – \; 25 \) — no constant term in the original, but the completed form still works.

Trap: \( x^2 + 5x + 25 \) — a student might match the 5 in the bracket directly to the coefficient of \( x \). But \( (x + 5)^2 \) expands to \( x^2 + 10x + 25 \), requiring a coefficient of \( 10x \). Since the coefficient here is \( 5x \), this cannot be written as \( (x + 5)^2 + q \). In fact, \( x^2 + 5x + 25 = (x + 2.5)^2 + 18.75 \).

Example: \( x^2 + 4x + 1 = (x + 2)^2 \; – \; 3 \)

Another: \( x^2 \; – \; 6x + 2 = (x \; – \; 3)^2 \; – \; 7 \)

Creative: \( x^2 + 2x = (x + 1)^2 \; – \; 1 \) — no constant term in the original, yet \( q \) is negative.

Trap: \( x^2 + 4x + 4 = (x + 2)^2 + 0 \) — this is a perfect square with \( q = 0 \), not negative. A student might assume every quadratic has a negative \( q \) value, but when \( c = (b/2)^2 \), the expression is already a perfect square and \( q = 0 \).

Example: \( x^2 + 2x + 1 = (x + 1)^2 \), minimum is 0

Another: \( x^2 \; – \; 10x + 25 = (x \; – \; 5)^2 \), minimum is 0

Creative: \( x^2 = (x + 0)^2 \), minimum is 0 — the simplest possible case, which students rarely think of.

Trap: \( x^2 + 2x \) — a student might think “there’s no constant, so the minimum is 0.” But \( x^2 + 2x = (x + 1)^2 \; – \; 1 \), so the minimum is \( -1 \), not 0. The “no constant means minimum is zero” misconception confuses the absence of a constant term with the expression having a minimum of zero.

Example: \( 2x^2 + 4x + 5 = 2(x + 1)^2 + 3 \), minimum is 3

Another: \( 2x^2 \; – \; 8x + 11 = 2(x \; – \; 2)^2 + 3 \), minimum is 3

Creative: \( 2x^2 + 3 = 2(x + 0)^2 + 3 \), minimum is 3 — no \( x \)-term at all.

Trap: \( 2x^2 + 4x + 3 \). A student might think, “half of 4 is 2, and the constant is 3, so the minimum is 3.” But because of the coefficient 2, we must factorize first: \( 2(x^2 + 2x) + 3 \). This becomes \( 2((x+1)^2 – 1) + 3 = 2(x+1)^2 – 2 + 3 = 2(x+1)^2 + 1 \). The minimum is actually 1.

Since \( q = c \; – \; (b/2)^2 \), whether \( q \) is positive depends on whether \( c \) is larger or smaller than \( (b/2)^2 \). When \( c > (b/2)^2 \), \( q \) is positive; when \( c < (b/2)^2 \), \( q \) is negative; when \( c = (b/2)^2 \), \( q = 0 \) and the expression is a perfect square. Students sometimes assume the completed square form always has a positive constant, perhaps because they associate squares with positive values.

True case: \( x^2 + 2x + 5 = (x + 1)^2 + 4 \), so \( q = 4 > 0 \). False case: \( x^2 + 6x + 2 = (x + 3)^2 \; – \; 7 \), so \( q = -7 < 0 \).

This is always true because \( (x + p)^2 \) expands to \( x^2 + 2px + p^2 \), so matching the coefficient of \( x \) gives \( 2p = b \), hence \( p = b/2 \). This works for positive, negative, odd, even, and fractional values of \( b \). For example, with \( x^2 \; – \; 6x + 1 \), \( b = -6 \), so \( p = -3 \) and we get \( (x \; – \; 3)^2 \). With \( x^2 + 5x + 1 \), \( b = 5 \), so \( p = 2.5 \) and we get \( (x + 2.5)^2 \).

The “use the full coefficient” misconception — putting \( b \) rather than \( b/2 \) inside the bracket — is one of the most common errors in completing the square.

The minimum value is \( c \; – \; (b/2)^2 \), which can be negative even when \( c \) is positive, provided \( (b/2)^2 > c \). This targets the “positive constant means positive minimum” misconception — students assume that because the original expression ends in a positive number, the expression can never be negative.

True case: \( x^2 + 2x + 5 = (x + 1)^2 + 4 \), minimum is \( 4 > 0 \). False case: \( x^2 + 6x + 5 = (x + 3)^2 \; – \; 4 \), minimum is \( -4 < 0 \), despite \( c = 5 \) being positive.

The turning point occurs at \( x = -b/2 \). When \( b > 0 \), we have \( -b/2 < 0 \), so the \( x \)-coordinate of the turning point is always negative. Students who make the “wrong sign for turning point” misconception read \( p = b/2 \) directly from the bracket and forget to reverse the sign, placing the turning point on the wrong side of the \( y \)-axis.

For example, \( y = x^2 + 6x + 10 \) has its turning point at \( x = -3 \), not \( x = 3 \). Note: the \( y \)-coordinate of the turning point can still be positive, negative, or zero — it’s the \( x \)-coordinate that is always negative when \( b > 0 \).

The student has made the “use the full coefficient” error. They placed 12 (the entire coefficient of \( x \)) inside the bracket instead of halving it. Expanding \( (x + 12)^2 \) gives \( x^2 + 24x + 144 \) — the coefficient of \( x \) is 24, not 12.

To get \( 12x \) in the expansion, we need \( (x + 6)^2 \), since \( (x + 6)^2 = x^2 + 12x + 36 \). The correct completed square form is \( (x + 6)^2 \; – \; 36 + 20 = (x + 6)^2 \; – \; 16 \).

The answer is correct — \( x^2 \; – \; 8x + 16 \) does equal \( (x \; – \; 4)^2 \) — but the reasoning reveals the “square root of the constant” misconception. The student found the bracket number by taking \( \sqrt{16} = 4 \), which only coincidentally matches the correct value of \( b/2 = 8/2 = 4 \).

The correct process is to halve the coefficient of \( x \): \( -8 \div 2 = -4 \), giving \( (x \; – \; 4)^2 \). This method fails on almost every other quadratic. For example, with \( x^2 \; – \; 8x + 20 \), the student’s method would give \( \sqrt{20} \approx 4.47 \) inside the bracket, but the correct answer is still \( (x \; – \; 4)^2 + 4 \). Getting the right answer with wrong reasoning gives false confidence that breaks down in harder problems.

The student correctly halved the coefficient of \( x \) but fell into the “forget to subtract \( (b/2)^2 \)” error. When we write \( (x + 5)^2 \), this expands to \( x^2 + 10x + 25 \) — it already contains a hidden \( +25 \). The student has effectively double-counted: the 25 is baked into the squared bracket, AND the 30 is tacked on separately.

The correct process is: \( x^2 + 10x + 30 = (x + 5)^2 \; – \; 25 + 30 = (x + 5)^2 + 5 \). The adjustment step (subtracting 25 then adding 30) is essential and the most commonly skipped part of the method.

The completing the square is correct — \( x^2 \; – \; 2x + 6 = (x \; – \; 1)^2 \; – \; 1 + 6 = (x \; – \; 1)^2 + 5 \) — but the student then made the “wrong sign for the turning point” error. From \( (x \; – \; 1)^2 + 5 \), the minimum occurs when \( x \; – \; 1 = 0 \), giving \( x = 1 \), not \( x = -1 \). The student read the \( -1 \) from the bracket and wrote that as the \( x \)-coordinate without changing the sign.

The correct turning point is \( (1, \; 5) \). Verify: when \( x = 1 \), \( y = 1 \; – \; 2 + 6 = 5 \). When \( x = -1 \), \( y = 1 + 2 + 6 = 9 \neq 5 \).

The student made the “forgetting to distribute the factor” error. When expanding \( 2[(x+2)^2 – 4] + 6 \), the factor of 2 must multiply everything inside the square brackets, including the subtracted 4. The student only multiplied the squared term by 2.

Correct steps: \( 2[(x+2)^2 – 4] + 6 = 2(x+2)^2 – 8 + 6 = 2(x+2)^2 – 2 \). The subtracted 4 becomes a subtracted 8 because of the 2 outside.