6.01f - Completing the square

Completing a monic square

Can you rewrite $x^2+6x+5$ so that the letter appears inside just one squared bracket?

Higher only. Start with the nearby square: $(x+3)^2=x^2+6x+9$. Its first two terms are exactly right, but its constant is four too large. Subtracting four repairs that difference without changing the expression.

The middle term in $(x+p)^2=x^2+2px+p^2$ is $2px$. Therefore halve the coefficient of $x$ to find the number inside the bracket. Add and subtract its square so the overall value stays unchanged:

For positive $x$, the area picture shows why adding $9$ completes the square. The algebraic identity also holds for zero and negative $x$, even though those are not positive side lengths.

[DIAGRAM: asset_name: Completing a square by adding the missing corner; asset_slug: completing-square-area; caption: Two strips of area 3x need a corner of area 9 to make a square of side x + 3.; description: A square is split into a blue x by x square, two sage rectangles each x by 3, and a clay 3 by 3 corner. Side segments are labelled x and 3. The total area is x squared plus 6x plus 9, equal to the square of x plus 3. This is an area model for positive x.]

Change only the sign of the linear term. The bracket number becomes negative, but its square is still positive.

The opening expression is therefore a square minus four. Unlike factorising into linear brackets, this method does not depend on finding integer factors of the quadratic.

Use expansion to check your own bracket and constant.

Completing squares with fractions

An odd linear coefficient produces a fraction inside the square, but the method is unchanged.

For example, half of $5$ is $5/2$, and $(5/2)^2=25/4$. Keeping these as fractions avoids rounding. Convert the original constant to quarters before combining it with $-25/4$.