Completing the Square: Free Response

The variable part is $x^2 - 9x$, so the coefficient to halve is $-9$, half of it is $-\tfrac{9}{2}$, and squaring that gives the completing constant:

$\left(-\tfrac{9}{2}\right)^2 = \tfrac{81}{4}.$

Add it and subtract it in the same line, so that the appearance of the expression changes and its value does not:

$x^2 - 9x + 4 = \left(x^2 - 9x + \tfrac{81}{4}\right) - \tfrac{81}{4} + 4.$

The bracket is the perfect square $\left(x - \tfrac{9}{2}\right)^2$, since half of $-9$ is $-\tfrac{9}{2}$ and that is the number the square carries. The loose constants combine over a denominator of $4$, writing $4$ as $\tfrac{16}{4}$:

$-\tfrac{81}{4} + \tfrac{16}{4} = -\tfrac{65}{4}.$

So the change of form is

$x^2 - 9x + 4 = \left(x - \tfrac{9}{2}\right)^2 - \tfrac{65}{4}.$

Check it by expanding. $\left(x - \tfrac{9}{2}\right)^2 = x^2 - 9x + \tfrac{81}{4}$, and subtracting $\tfrac{65}{4}$ leaves $x^2 - 9x + \tfrac{16}{4}$, which is $x^2 - 9x + 4$. The two expressions agree for every $x$, which is what makes this a rewrite rather than an approximation.

Nothing is distributed or expanded here. Put the completed form in place of the left-hand side:

$\left(x - \tfrac{9}{2}\right)^2 - \tfrac{65}{4} = 0.$

Isolate the square by adding $\tfrac{65}{4}$ to both sides:

$\left(x - \tfrac{9}{2}\right)^2 = \tfrac{65}{4}.$

The right-hand side is positive, so the root of both sides carries a plus-or-minus. Writing only the positive root would quietly throw the second solution away:

$x - \tfrac{9}{2} = \pm\tfrac{\sqrt{65}}{2}.$

The denominator $4$ is a perfect square, so its root is $2$; the numerator $65 = 5 \cdot 13$ is not, so $\sqrt{65}$ does not simplify and the answer stays exact rather than decimal. Adding $\tfrac{9}{2}$ to both sides:

$x = \tfrac{9}{2} \pm \tfrac{\sqrt{65}}{2} = \dfrac{9 \pm \sqrt{65}}{2}.$

Check the first one. With $x = \dfrac{9 + \sqrt{65}}{2}$,

$x^2 = \dfrac{81 + 18\sqrt{65} + 65}{4} = \dfrac{146 + 18\sqrt{65}}{4},$

while $-9x = \dfrac{-162 - 18\sqrt{65}}{4}$ and $4 = \dfrac{16}{4}$. The two terms carrying $\sqrt{65}$ cancel exactly, and the plain numbers give $146 - 162 + 16 = 0$, so the total is $0$ and the value satisfies the equation.

Count the places $x$ appears. In $x^2 - 9x + 4$ it appears in two terms, of different degree, and there is no way to gather them: there is no single $x$ to solve for, and no way to see how the two terms behave against each other as $x$ moves.

The completed form has $x$ in exactly one place, inside the square:

$x^2 - 9x + 4 = \left(x - \tfrac{9}{2}\right)^2 - \tfrac{65}{4}.$

That single occurrence is what both readings exploit, in opposite directions.

Read the form as it stands. A real square is never negative, so $\left(x - \tfrac{9}{2}\right)^2 \ge 0$, with equality exactly when $x = \tfrac{9}{2}$. The whole expression is therefore at least $-\tfrac{65}{4}$, and equals it only there. So the vertex is $\left(\tfrac{9}{2},\ -\tfrac{65}{4}\right)$, the axis of symmetry is $x = \tfrac{9}{2}$, and the smallest value the expression takes is $-\tfrac{65}{4}$. No equation was solved to get any of that.

Now set the same form equal to zero instead. Because $x$ occurs once, the square can be moved onto one side by itself, and undoing a square is a move that exists: take the root of both sides and attach a plus-or-minus. That move is simply not available while $x$ sits in two terms at once, which is why the standard form has to be left behind first.

So the two payoffs are not two techniques that happen to share a first step. They are one structural fact, that every $x$ has been collected into a single square, used first for reading and then for solving.

Factor $-4$ out of the $x^2$ and $x$ terms only. The loose constant carries no $x$, so it stays outside the bracket:

$-4x^2 + 10x + 1 = -4\left(x^2 - \tfrac{5}{2}x\right) + 1.$

Dividing $10$ by $-4$ gives $-\tfrac{5}{2}$, and the sign is worth a pause: pulling out a negative flips the sign of the middle term inside the bracket.

Inside, the coefficient of $x$ is $-\tfrac{5}{2}$, half of it is $-\tfrac{5}{4}$, and squaring gives $\tfrac{25}{16}$. Add and subtract that inside the bracket:

$-4\left(x^2 - \tfrac{5}{2}x + \tfrac{25}{16} - \tfrac{25}{16}\right) + 1 = -4\left(\left(x - \tfrac{5}{4}\right)^2 - \tfrac{25}{16}\right) + 1.$

Now the step the whole method turns on. The $-4$ multiplies everything the bracket holds, the subtracted constant included:

$-4 \cdot \left(-\tfrac{25}{16}\right) = \tfrac{25}{4},$

so the finished form is

$g(x) = -4\left(x - \tfrac{5}{4}\right)^2 + \tfrac{25}{4} + 1 = -4\left(x - \tfrac{5}{4}\right)^2 + \tfrac{29}{4}.$

Expand back to check. $-4\left(x^2 - \tfrac{5}{2}x + \tfrac{25}{16}\right) = -4x^2 + 10x - \tfrac{25}{4}$, and $-\tfrac{25}{4} + \tfrac{29}{4} = \tfrac{4}{4} = 1$, which returns $-4x^2 + 10x + 1$.

Read the finished form against the shape $a(x - h)^2 + k$. Here $a = -4$, $h = \tfrac{5}{4}$ and $k = \tfrac{29}{4}$, so the vertex is $\left(\tfrac{5}{4},\ \tfrac{29}{4}\right)$ and the axis of symmetry is the vertical line $x = \tfrac{5}{4}$.

Which kind of extreme point that vertex is follows from two facts and no testing. First, a real square is never negative:

$\left(x - \tfrac{5}{4}\right)^2 \ge 0 \quad \text{for every real } x.$

Second, multiplying both sides by the negative number $-4$ reverses the comparison:

$-4\left(x - \tfrac{5}{4}\right)^2 \le 0.$

Adding $\tfrac{29}{4}$ to both sides gives $g(x) \le \tfrac{29}{4}$ for every real $x$. Equality needs the square to be zero, which happens only at $x = \tfrac{5}{4}$. So the vertex is the highest point of the graph, and the maximum value is $\tfrac{29}{4}$, attained exactly once.

That last word decides the range. The value $\tfrac{29}{4}$ is reached, so it belongs to the range; and the square can be made as large as you like by taking $x$ far from $\tfrac{5}{4}$, so $-4$ times it runs down without end. The outputs therefore fill everything from $\tfrac{29}{4}$ downwards: the range is $g(x) \le \tfrac{29}{4}$.

Notice that the sign of the leading coefficient did two jobs at once. It turned the comparison around, which made the vertex a maximum, and it is the same sign that opens the parabola downward.

Work down the lines and stop at the first that fails.

Line 1 is correct: the $5$ comes out of the $x^2$ and $x$ terms and the loose constant stays outside. Line 2 is correct too: inside the bracket the coefficient of $x$ is $-4$, half of it is $-2$, squaring gives $4$, and adding and subtracting that turns $x^2 - 4x$ into $(x - 2)^2 - 4$.

Line 3 is the first line that fails. The $5$ outside the bracket multiplies everything inside it, and the bracket holds two things:

$5\left((x - 2)^2 - 4\right) = 5(x - 2)^2 - 20.$

The student multiplied the square by $5$ and carried the other piece out untouched. Written correctly the line reads

$5(x - 2)^2 - 20 + 3 = 5(x - 2)^2 - 17,$

so the smallest value of the expression is $-17$, taken at $x = 2$, not $-1$.

Two checks expose this without running the method again. The first is to expand the student's form:

$5(x - 2)^2 - 1 = 5x^2 - 20x + 20 - 1 = 5x^2 - 20x + 19,$

which is not the expression that was handed in, so the rewrite cannot be an identity. The second is to evaluate both versions at a single convenient input. At $x = 2$ the original gives

$5(4) - 20(2) + 3 = 20 - 40 + 3 = -17,$

while the student's form gives $5(0) - 1 = -1$. One input is enough to convict, because a correct rewrite has to agree everywhere, so a single disagreement settles it.

Name the sides before writing anything down. Two sides run away from the wall, each of length $w$ metres, and one side runs along the wall; the wall itself is not fenced. So the fencing is spent on three sides:

$2w + \ell = 46,$

where $\ell$ is the length in metres of the side along the wall. That single equation in two unknowns is exactly what lets one of them be eliminated:

$\ell = 46 - 2w.$

The area is the product of the two dimensions, so substituting for $\ell$ leaves a function of $w$ alone:

$A(w) = w\ell = w(46 - 2w) = -2w^2 + 46w,$

measured in square metres.

Not every number is a possible $w$. A side of length zero or less is not a side, so $w > 0$; and the side along the wall must also be positive, which needs $46 - 2w > 0$, that is $w < 23$. So the plots that can actually be built are those whose $w$ lies strictly between $0$ and $23$ metres. The restriction comes from the situation, not from the algebra, which would happily accept any number at all.

Complete the square on $A(w) = -2w^2 + 46w$. Both terms carry a $w$, so factor $-2$ out of both:

$A(w) = -2\left(w^2 - 23w\right).$

Inside, half of $-23$ is $-\tfrac{23}{2}$ and squaring gives $\tfrac{529}{4}$, so add and subtract that inside the bracket:

$A(w) = -2\left(\left(w - \tfrac{23}{2}\right)^2 - \tfrac{529}{4}\right).$

Distribute the $-2$ across both pieces, the subtracted constant included:

$A(w) = -2\left(w - \tfrac{23}{2}\right)^2 + \tfrac{529}{2}.$

Now read it. The square is never negative and $-2$ is negative, so the first term is at most zero and therefore

$A(w) \le \tfrac{529}{2} \quad \text{for every } w,$

with equality exactly when the square is zero, at $w = \tfrac{23}{2} = 11.5$. That width lies inside the permitted range, so it is a plot the club can really build, which had to be checked rather than assumed.

The largest area is therefore $\tfrac{529}{2} = 264.5$ square metres. The dimensions are $11.5$ metres for each of the two sides running from the wall and $46 - 2(11.5) = 23$ metres along the wall. Check both records: $11.5 \times 23 = 264.5$ square metres, and the fencing used is $11.5 + 11.5 + 23 = 46$ metres, all of it.

Set the completed form equal to the target:

$-2\left(w - \tfrac{23}{2}\right)^2 + \tfrac{529}{2} = 240.$

Move the constant across and divide by $-2$, so that the square is standing alone. Since $\tfrac{529}{2} - 240 = \tfrac{529 - 480}{2} = \tfrac{49}{2}$,

$-2\left(w - \tfrac{23}{2}\right)^2 = -\tfrac{49}{2}, \qquad \left(w - \tfrac{23}{2}\right)^2 = \tfrac{49}{4}.$

The right-hand side is positive, so take the root of both sides and attach the plus-or-minus:

$w - \tfrac{23}{2} = \pm\tfrac{7}{2}.$

That gives $w = \tfrac{23}{2} + \tfrac{7}{2} = 15$ and $w = \tfrac{23}{2} - \tfrac{7}{2} = 8$.

Both lie strictly between $0$ and $23$, so both describe plots that can be built, and the two plots are genuinely different rather than the same one written twice. At $w = 8$ the side along the wall is $46 - 16 = 30$ metres, and $8 \times 30 = 240$ square metres. At $w = 15$ it is $46 - 30 = 16$ metres, and $15 \times 16 = 240$ square metres. Each uses $8 + 8 + 30 = 46$ metres and $15 + 15 + 16 = 46$ metres of fencing respectively, all of it.

So the club really does have two choices, and nothing here rules a value out. A negative width would have been rejected, and so would a width big enough to leave no wall side, but neither happens: in this situation any target area strictly between zero and the greatest available is met at two different widths, one on each side of the width that gives the greatest area, and here both of them fit.

Two routes, and they agree.

The first needs no equation at all. Part B established that the area never exceeds $\tfrac{529}{2} = 264.5$ square metres, and $264.5$ is less than $280$. A quantity that never exceeds $264.5$ cannot equal $280$. That is the entire argument, and it is available only because the completed form states the ceiling outright; the standard form $-2w^2 + 46w$ states no ceiling anywhere.

The second route runs the algebra anyway, because what it produces is the point of the part. Setting the area equal to the target,

$-2\left(w - \tfrac{23}{2}\right)^2 + \tfrac{529}{2} = 280.$

Writing $280$ as $\tfrac{560}{2}$ and moving the constant across,

$-2\left(w - \tfrac{23}{2}\right)^2 = \tfrac{560 - 529}{2} = \tfrac{31}{2},$

and dividing by $-2$ isolates the square:

$\left(w - \tfrac{23}{2}\right)^2 = -\tfrac{31}{4}.$

A real square is never negative, and the right-hand side is negative, so no real number $w$ satisfies this. There is nothing further to do and nothing has gone wrong. The method finished, and what it delivered is the information that the target cannot be met.

That is the point worth keeping. An isolated square equal to a negative number is an answer, not a stuck calculation, and here it says something concrete about fencing: no rectangular plot of this kind can enclose $280$ square metres with $46$ metres of fence, whatever its dimensions. The club must either accept at most $264.5$ square metres, build something that is not a rectangle, or buy more fencing.

This is an equation, not a bare expression, so it has a second side and may be divided through by the leading coefficient. Every term goes:

$x^2 + \tfrac{5}{2}x - 1 = 0.$

Move the constant to the right, leaving the variable terms alone and ready to be completed:

$x^2 + \tfrac{5}{2}x = 1.$

Half of $\tfrac{5}{2}$ is $\tfrac{5}{4}$, and squaring gives $\tfrac{25}{16}$. Add it to BOTH sides, which is the equation's version of adding and subtracting in the same line:

$\left(x + \tfrac{5}{4}\right)^2 = 1 + \tfrac{25}{16} = \tfrac{41}{16}.$

The right-hand side is positive, so there are two real solutions and the root carries a plus-or-minus. The denominator $16$ is a perfect square while $41$ is prime, so

$x + \tfrac{5}{4} = \pm\tfrac{\sqrt{41}}{4}, \qquad x = \dfrac{-5 \pm \sqrt{41}}{4}.$

Check the first. With $x = \dfrac{-5 + \sqrt{41}}{4}$,

$x^2 = \dfrac{25 - 10\sqrt{41} + 41}{16} = \dfrac{33 - 5\sqrt{41}}{8},$

so $2x^2 = \dfrac{33 - 5\sqrt{41}}{4}$, while $5x = \dfrac{-25 + 5\sqrt{41}}{4}$. The terms carrying the root cancel and $\dfrac{33 - 25}{4} = 2$, so $2x^2 + 5x - 2 = 2 - 2 = 0$.

Start with what each route promises, because that is where they differ, not in how hard the arithmetic is.

An integer factor search promises a finite check and nothing more. For $2x^2 + 5x - 2$ you look for integers filling $(2x + m)(x + n)$, which needs $mn = -2$ and $2n + m = 5$. There are four integer pairs with product $-2$, and the middle coefficients they produce are

$-3, \quad 3, \quad 0, \quad 0,$

none of which is $5$. The list is exhausted and the search stops exactly there. It has established something real, that no integer pair works, but that is not a solution and does not point at one. Its promise was only ever to test candidates, so when every candidate fails it has nothing left to say about the roots.

Completing the square makes no such bargain, because it never asks a question that could be answered 'no candidates left'. Every step is a computation rather than a search: halve the coefficient of $x$, square the result, add and subtract it, and tidy up. Halving a real number always returns a real number, and squaring one always returns a real number, so the finished form is reached whatever the coefficients are, this quadratic included.

The two therefore fail in incomparable ways. One can run out of candidates; the other has no candidates to run out of. That is the sense in which completing the square never fails, and it is why this lesson sits after factoring rather than beside it.

The answer from part A shows the gap concretely. Its two numbers are irrational, and any factorization into linear factors with integer coefficients would have forced them to be rational, so they were out of the factor search's reach before it started.

Take the claim apart. Its premise is true and its conclusion does not follow from it.

What never fails is the rewrite. Given any $ax^2 + bx + c$ with $a \ne 0$, every step of the method is a computation, so the steps always finish, and they always finish in the same shape: one square, times a constant, plus a constant. That much is guaranteed in advance.

What the finished form then says is a separate question, and the form is what answers it. Isolate the square and the equation reads

$(x - h)^2 = k$

for some number $k$ that the coefficients decide. Exactly one of three things is true of $k$. If $k > 0$ there are two real solutions, $x = h \pm \sqrt{k}$. If $k = 0$ those two collapse into the single solution $x = h$. If $k < 0$ there is no real solution at all, because a real square is never negative.

So the method is a reporter, not a guarantor. It always reaches the crossroads; it never promises which way the crossroads points.

The supporting equation makes that concrete. For $x^2 + 5x + 9$, half of $5$ is $\tfrac{5}{2}$ and squaring gives $\tfrac{25}{4}$, so

$x^2 + 5x + 9 = \left(x + \tfrac{5}{2}\right)^2 - \tfrac{25}{4} + 9 = \left(x + \tfrac{5}{2}\right)^2 + \tfrac{11}{4}.$

Setting that equal to zero and isolating the square,

$\left(x + \tfrac{5}{2}\right)^2 = -\tfrac{11}{4},$

which no real $x$ satisfies. The method did not fail on this equation; it succeeded, and what it produced is the information that the equation has no real solution. Read the very same form the other way and it says the smallest value of $x^2 + 5x + 9$ is $\tfrac{11}{4}$, comfortably above zero, which is the same fact seen from the other side: the graph sits entirely above the axis, so there is nothing for it to cross.

The repair to the claim is to say precisely what never fails. Reaching the completed form never fails. Finding a real root is not something any method could promise, because for some equations there is nothing there to find.

The claim is an 'if and only if', so there are two things to prove and they are not the same thing.

First direction, that the condition is necessary. Suppose $x^2 + bx + d = (x + p)^2$ for some real $p$ and for every $x$. Expanding the right-hand side,

$x^2 + bx + d = x^2 + 2px + p^2.$

Two polynomials agree at every $x$ only if their matching coefficients agree, so the $x$ terms give $b = 2p$ and the constants give $d = p^2$. (If you would rather not appeal to that principle, substitute $x = 0$ to get $d = p^2$, then $x = 1$ to get $1 + b + d = 1 + 2p + p^2$, which with $d = p^2$ leaves $b = 2p$.) The first equation forces $p = \tfrac{b}{2}$, and substituting into the second gives

$d = p^2 = \left(\tfrac{b}{2}\right)^2.$

So no other constant is available: this direction rules every other value of $d$ out.

Second direction, that the condition is sufficient. Suppose $d = \left(\tfrac{b}{2}\right)^2$. Take $p = \tfrac{b}{2}$ and expand:

$\left(x + \tfrac{b}{2}\right)^2 = x^2 + 2 \cdot \tfrac{b}{2} \cdot x + \left(\tfrac{b}{2}\right)^2 = x^2 + bx + d.$

So such a $p$ exists whenever the condition holds. The two directions together make the condition exact rather than merely one that happens to work.

How many such $p$ can there be? The first direction pinned it down with no choice left: $2p = b$ is a linear equation with exactly one solution. So $p$ is unique, and the familiar instruction to halve the coefficient of $x$ is not one recipe among several but the only one there is, which is why the method needs no guessing anywhere.

One consequence is worth noticing on the way out. Since $d$ must equal $\left(\tfrac{b}{2}\right)^2$, and a real square is never negative, no trinomial $x^2 + bx + d$ with $d < 0$ is a perfect square, whatever $b$ may be.