IGCSE Maths / Quadratic equations, curves and inequalities
Completing the square with a negative coefficient revision notes
Edexcel International GCSE Maths (4MA1), chapter 22: Quadratic equations, curves and inequalities, specification point 2.2d. Free lesson.
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Completing the square when the x² term is negative means taking −1 out of the x terms first, then completing the square inside the bracket. The completed form also shows the turning point, which is a maximum when the coefficient of the squared bracket is negative.
Take −1 out of the x terms first
In −x² + 6x − 1, only −x² and 6x contain x. Taking −1 out of those terms leaves x² − 6x inside the bracket. The constant −1 stays outside because it is not part of the x terms. −x² + 6x − 1 = −(x² − 6x) − 1
Complete the square inside the bracket
Now use the usual completing-the-square method on x² − 6x. Half of −6 is −3, so the squared bracket is (x − 3)². Squaring this adds 9, so subtract 9 inside the bracket to keep the expression unchanged.
Multiply back out and keep the signs
The −1 outside multiplies everything inside the bracket. In particular, −1 × −9 = +9. Then collect the constants to get the completed-square form. You can check it by expanding: it returns the original expression. The completed form is −(x − 3)² + 8. Check: −x² + 6x − 9 + 8 = −x² + 6x − 1.
Read the maximum point from the completed form
The completed form shows the turning point directly. In −(x − 3)² + 8, the squared part is zero when x = 3, so y = 8 there. Since the coefficient of the squared bracket is negative, the curve opens downwards and this turning point is a maximum, not a minimum. The turning point is (3, 8). A negative coefficient in front of the squared bracket means the curve has a maximum.
Use the same method with another quadratic
For −x² + 8x − 3, take −1 out of the x terms, complete the square inside, then multiply back and collect the constants. The completed form is −(x − 4)² + 13.