AL5 - Completing the square

Building a complete square

How can you know that $x^2-10x+37$ never falls below $12$, without trying lots of values of $x$? The expanded expression hides that lower limit. A squared bracket will make it visible.

Start with $x^2+6x$. For positive $x$, think of a square of side $x$ and two strips, each of width $3$. Together they almost make a larger square of side $x+3$, but its corner of area $9$ is missing. Adding $9$ completes that square.

[DIAGRAM: asset_name: Completing a square with two strips; asset_slug: completing-square-area; caption: Two strips of area 3x leave a corner of area 9 to complete a square; description: A square with total side length x plus 3 is partitioned into a blue x by x square labelled x squared, two sage rectangles each labelled 3x, and a clay 3 by 3 corner labelled 9. Side labels show x and 3 along the bottom and left. A sentence below states that x squared plus 6x plus 9 equals x plus 3 all squared, for positive x in the area picture.]

The picture uses positive lengths, but expansion gives the identity $x^2+6x+9\equiv(x+3)^2$ for every real $x$. If the expression starts with $x^2+6x+2$, adding $9$ would change it. Instead, put in the square and subtract the extra $9$: $x^2+6x+2\equiv(x+3)^2-9+2$.

The coefficient of $x$ tells you the number inside the bracket: halve it, because expansion doubles that number in the middle term.

Use that compensation explicitly before collecting the constant terms.

Now change only the sign of the middle term. The bracket number changes sign, but its square remains positive.

You can check every completed square by expanding it back; this catches both a wrong bracket and a wrong constant.

Keep the coefficient and the bracket number separate in the next calculation.

Leading coefficients and fractional shifts

When the coefficient of $x^2$ is not $1$, first factor it out of the two variable terms so that you complete a monic quadratic inside the bracket.

In $2x^2+12x+5$, write $2(x^2+6x)+5$. The number to halve is now $6$, not $12$. Any correction made inside the bracket is also multiplied by $2$ outside it.

The general identity describes the result, but the short bracket-by-bracket method keeps the correction visible.

Now change only the linear coefficient, from $12$ to $10$. After factoring out $2$, the inner coefficient is odd, so halving it produces a fraction. Keep that fraction exact.

Fractions are also behind a neat numerical fact. The product of two consecutive integers is exactly $\frac14$ below the square of their halfway point: $20\times21=420$, while $20.5^2=420.25$. Algebra explains every such case:

The identity is not on the formulae sheet: it is another completed square, with $b=1$ and $c=0$.

Try a coefficient that produces a fraction after the first factorisation.

Reading greatest and least values

A completed square isolates the only part that varies, so its sign tells you whether the constant outside is a lower or an upper limit.

For a real input, $(x+p)^2\ge0$, with equality exactly when $x=-p$. If $a>0$, then $a(x+p)^2+q\ge q$: its least value is $q$. If $a<0$, multiplying by $a$ reverses the comparison, so $a(x+p)^2+q\le q$: its greatest value is $q$. In both cases that value occurs when the bracket is zero. These facts are not on the formulae sheet: learn them.

The sign inside the bracket is opposite to the input that makes it zero. A form containing $(x-2)^2$ reaches its limiting value at $x=2$.

Now reverse the sign of the two variable terms, leaving the constant fixed. The square has a negative multiplier, so the same idea gives a greatest value instead.

We can now settle the opening question in one line: $x^2-10x+37\equiv(x-5)^2+12$. The square cannot be negative, so the expression cannot fall below $12$, and it equals $12$ at $x=5$. Completing the square has turned a claim about infinitely many inputs into a statement about one square.

Read the sign outside the square to decide greatest or least, then solve the bracket equal to zero to locate it. The two signs do different jobs.

Handle both a negative multiplier and a fractional shift in the final question.

Quick Recap

• Complete a monic quadratic by halving the linear coefficient, squaring that half, and compensating for the introduced constant.
• For a non-monic quadratic, factor the leading coefficient from the variable terms first; it multiplies the correction too.
• Fractions inside the bracket are normal. Keep them exact and check by expanding back.
• In $a(x+p)^2+q$, the limiting value is $q$, at $x=-p$: a least value if $a>0$, a greatest value if $a<0$.

The opening expression is $(x-5)^2+12$, so its least value is exactly $12$, attained at $x=5$.