Why does completing the square on x squared plus 8x minus 5 give (x+4) squared minus 21, not (x+4) squared minus 5?

Because the bracket (x+4) squared expands to x squared plus 8x plus 16, which is 16 more than x squared plus 8x. That extra 16 must be subtracted to keep the expression equal to the original. So the constant becomes minus 5 minus 16, which equals minus 21, not minus 5.

What is the completing-the-square rule for the constant?

The constant in the completed-square form equals the original constant minus a squared, where a is half the coefficient of x. For x squared plus 8x minus 5, a equals 4 and a squared equals 16, so the constant is minus 5 minus 16 equals minus 21.

How do you complete the square when the coefficient of x squared is not 1?

Factor out the leading coefficient from the x squared and x terms first. For 2x squared minus 12x plus 7, write 2 times (x squared minus 6x) plus 7. Then complete the square inside: x squared minus 6x equals (x minus 3) squared minus 9. Substitute back and distribute: 2 times ((x minus 3) squared minus 9) plus 7 equals 2(x minus 3) squared minus 18 plus 7 equals 2(x minus 3) squared minus 11.

GCSE Maths Higher · AQA · Completing the square and turning points

Completing the square: forgetting to subtract the adjustment constant

When students complete the square on $x^2+8x-5$ , the instinct is to write the bracket correctly as $(x+4)^2$ and then copy the original constant unchanged, giving $(x+4)^2-5$ . That answer is wrong. The bracket $(x+4)^2$ expands to $x^2+8x+16$ , which is 16 more than $x^2+8x$ . That hidden +16 must be subtracted, so the correct constant is $-5-16=-21$ and the answer is $(x+4)^2-21$ .

The rule is: constant in the answer = original constant $-\,a^2$ , where $a$ is half the coefficient of $x$ . Missing this step is one of the most reliable grade 7 to 9 mark-losers on AQA Higher papers.

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How to spot it in your own work

You wrote $(x+4)^2-5$ for $x^2+8x-5$ , keeping the original constant $-5$ instead of subtracting $a^2=16$ to get $-21$ .
You wrote $(x+3)^2+13$ for $x^2+6x+13$ instead of $(x+3)^2+4$ , failing to subtract $a^2=9$ from $+13$ .
You wrote $(x-9)^2+70$ for $x^2-18x+70$ instead of $(x-9)^2-11$ , keeping $+70$ when you should subtract $a^2=81$ .
For a leading coefficient other than 1, you forgot to scale the adjustment: writing $2(x-3)^2-2$ for $2x^2-12x+7$ instead of $2(x-3)^2-11$ , because $2\times9=18$ not $9$ .

An exam question that triggers it

Here is the structure of a typical AQA Higher completing-the-square question:

Write $x^2+8x-5$ in the form $(x+p)^2+q$ .

State the values of $p$ and $q$ .

The misconception writes $(x+4)^2-5$ (keeps the original constant). The fix: $(x+4)^2$ expands to $x^2+8x+16$ , so subtract 16 and carry $-5$ : $-5-16=-21$ , giving $(x+4)^2-21$ .

Why students fall for this

The bracket step is the visible, satisfying part of completing the square. Students write $(x+4)^2$ , feel they are done, and copy the original constant unchanged. The hidden arithmetic $(x+4)^2 = x^2+8x+16$ is out of view at the moment the constant is written down.

A second error pattern appears when the original constant is positive. Students complete the square correctly for negative constants (where the magnitude grows from $-5$ to $-21$ ) but hesitate when the constant should reduce: for $x^2+6x+13$ , the constant goes from $+13$ to $+4$ , which feels counterintuitive. They may write $(x+3)^2-4$ (wrong subtraction order) or keep $+13$ .

For quadratics with a leading coefficient other than 1, both errors compound: students who correctly factor out the coefficient often forget to scale the adjustment by that coefficient when distributing back.

The fix: Always subtract a² from the original constant: constant in answer = original constant − a²

Step 1: find a. Halve the coefficient of $x$ . For $x^2+8x-5$ , the coefficient of $x$ is 8, so $a = 4$ .

Step 2: write the bracket. $(x+a)^2 = (x+4)^2$ .

Step 3: compute the adjustment. $a^2 = 16$ . The constant in the answer is $\text{original constant} - a^2 = -5 - 16 = -21$ .

Step 4: write the answer. $x^2+8x-5 = (x+4)^2 - 21$ .

Step 5: verify by expanding. Expand $(x+4)^2 - 21 = x^2+8x+16-21 = x^2+8x-5$ . If the expansion does not match the original, there is an arithmetic error in step 3.

Worked example

Write $x^2+8x-5$ in the form $(x+p)^2+q$ .

Find a. Half of 8 is 4, so $a=4$ , bracket $(x+4)^2$ .
Compute adjustment. $a^2 = 16$ . Constant $= -5 - 16 = -21$ .
Answer.
$x^2+8x-5 = (x+4)^2 - 21$
Trap: $(x+4)^2-5$ expands to $x^2+8x+11$ , wrong.
Verify. $(x+4)^2-21 = x^2+8x+16-21 = x^2+8x-5$ ✓

Write $2x^2-12x+7$ in the form $2(x+p)^2+q$ .

Factor out 2 from the x terms. $2x^2-12x+7 = 2(x^2-6x)+7$ .
Complete the square inside. $x^2-6x = (x-3)^2-9$ (a=3, a²=9).
Substitute and distribute.
$2((x-3)^2-9)+7 = 2(x-3)^2 - 18 + 7 = 2(x-3)^2-11$
Trap: forgetting to scale gives $2(x-3)^2-9+7 = 2(x-3)^2-2$ , wrong.
Verify. $2(x-3)^2-11 = 2(x^2-6x+9)-11 = 2x^2-12x+18-11 = 2x^2-12x+7$ ✓

Show that $x^2+6x+13$ is always positive.

Complete the square. $x^2+6x+13 = (x+3)^2+4$ (a=3, a²=9, constant = 13−9 = 4).
Prove positivity. $(x+3)^2 \geq 0$ for all real $x$ , so
$(x+3)^2 + 4 \geq 4 > 0$
The minimum value is 4, achieved when $x=-3$ .

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Common questions

Why is the answer $(x+4)^2-21$ and not $(x+4)^2-5$ for $x^2+8x-5$ ?: Because $(x+4)^2 = x^2+8x+16$ , which is 16 more than $x^2+8x$ . To keep the expression equal to $x^2+8x-5$ you must subtract that extra 16. The constant becomes $-5-16=-21$ . Keeping $-5$ unchanged gives $(x+4)^2-5 = x^2+8x+11$ , which is a different expression.
How does the rule change when the constant in $x^2+6x+13$ is positive?: The rule is the same: subtract $a^2$ from the original constant. Here $a=3$ and $a^2=9$ , so the constant becomes $13-9=4$ and the answer is $(x+3)^2+4$ . The positive constant reduces (from 13 to 4) rather than becoming more negative, but the same subtraction rule applies.
How do you complete the square when the coefficient of $x^2$ is not 1?: Factor out the leading coefficient from the $x^2$ and $x$ terms first. For $2x^2-12x+7$ , write $2(x^2-6x)+7$ , then complete the square inside: $(x-3)^2-9$ . Distribute the 2 back: $2(x-3)^2-18+7 = 2(x-3)^2-11$ . The adjustment $a^2=9$ must be scaled by the leading coefficient 2 to give 18, not left as 9.

Related misconceptions

Turning point x-coordinate: the sign flips from vertex formCopying the printed sign from (x−7)²+8 to get x=−7 instead of solving (x−7)=0 to get x=7, a companion completing-the-square misconception on the same Higher topic.
Negative indices mean reciprocals, not negative numbersReading the minus in 8^-5 as a sign on the result and writing −8^5, when the correct value is 1/8^5, a different Higher topic with the same trap structure of misreading a sign.

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