Completing the Square: Free Response

No equation here, so no second side to balance: add the completing constant and subtract it in the same line, which changes nothing but lets you regroup.

The linear coefficient is $b = -10$, so half is $-5$ and the completing constant is $(-5)^2 = 25$:

$x^2 - 10x + 7 = \left(x^2 - 10x + 25\right) - 25 + 7.$

The bracket is a perfect square; the loose constants combine to $-25 + 7 = -18$:

$x^2 - 10x + 7 = (x - 5)^2 - 18.$

Expanding back checks it:

$(x - 5)^2 - 18 = x^2 - 10x + 25 - 18 = x^2 - 10x + 7.$

Move the constant to the right so the left side holds only the $x$ terms:

$x^2 + 5x = -2.$

Here $b = 5$ is odd, so half is the fraction $\tfrac{5}{2}$ and the completing constant is $\left(\tfrac{5}{2}\right)^2 = \tfrac{25}{4}$. Add it to BOTH sides, writing $-2$ as $-\tfrac{8}{4}$ so the right side combines:

$x^2 + 5x + \frac{25}{4} = -\frac{8}{4} + \frac{25}{4} = \frac{17}{4}.$

The left side is now the square of $x + \tfrac{5}{2}$. Take the square root of both sides, keeping BOTH signs, and use $\sqrt{\tfrac{17}{4}} = \tfrac{\sqrt{17}}{2}$, since the denominator is a perfect square:

$\left(x + \frac{5}{2}\right)^2 = \frac{17}{4} \quad\Rightarrow\quad x + \frac{5}{2} = \pm\frac{\sqrt{17}}{2}.$

Subtract $\tfrac{5}{2}$ and combine over the common denominator $2$:

$x = -\frac{5}{2} \pm \frac{\sqrt{17}}{2} = \frac{-5 \pm \sqrt{17}}{2}.$

Start from the pattern. Squaring a binomial always produces

$(x + p)^2 = x^2 + 2px + p^2.$

For $x^2 + bx + (\text{constant})$ to BE such a square, the middle terms must agree: $2px = bx$. That fixes $p$ with no choice left,

$2p = b \quad\Rightarrow\quad p = \frac{b}{2},$

and once $p$ is pinned down the last term of the square gives the constant:

$\text{constant} = p^2 = \left(\frac{b}{2}\right)^2.$

So the halving is forced by the $2$ in $2px$, itself produced because squaring $x + p$ generates the cross term twice, once from each bracket.

The sign takes care of itself too. If $b$ is negative, $p = \tfrac{b}{2}$ is negative, the binomial reads $x - \left|\tfrac{b}{2}\right|$, and $p^2$ is still positive, as with the $-10$ in part A.

The $x^2$ term does not stand alone, so nothing can complete yet. This is an EQUATION with right side $0$, so divide every term by the leading coefficient $5$ (dividing $0$ by $5$ leaves $0$):

$x^2 + 4x - 3 = 0.$

Move the constant across, halve $b = 4$ to get $2$, and add $2^2 = 4$ to both sides:

$x^2 + 4x + 4 = 3 + 4 = 7 \quad\Rightarrow\quad (x + 2)^2 = 7.$

Take the square root of both sides, keeping both signs, and isolate $x$:

$x + 2 = \pm\sqrt{7} \quad\Rightarrow\quad x = -2 \pm \sqrt{7}.$

Since $\sqrt{7}$ is between $2$ and $3$, the roots are roughly $0.65$ and $-4.65$, straddling $-2$ as a quick check.

Dividing an expression by $3$ gives a DIFFERENT expression, one third as large. No equals sign means nothing to balance, so the leading coefficient must be factored out, not divided away.

Factor the $3$ out of the two $x$ terms, leaving the constant alone:

$3x^2 - 18x + 32 = 3\left(x^2 - 6x\right) + 32.$

Complete the square inside the bracket. Half of $-6$ is $-3$, and $(-3)^2 = 9$, so add and subtract $9$ INSIDE it:

$3\left(x^2 - 6x + 9 - 9\right) + 32 = 3\left[(x - 3)^2 - 9\right] + 32.$

Distribute the $3$ and combine constants:

$3(x - 3)^2 - 27 + 32 = 3(x - 3)^2 + 5.$

Expanding back confirms it: $3\left(x^2 - 6x + 9\right) + 5 = 3x^2 - 18x + 27 + 5 = 3x^2 - 18x + 32$.

The completed form mentions $x$ exactly once, inside a square, whose sign you already know.

For every real $x$, $(x - 3)^2$ is never negative, so multiplying by the positive $3$ keeps it that way:

$3(x - 3)^2 \ge 0.$

Add $5$ to both sides and the expression is bounded below:

$3(x - 3)^2 + 5 \ge 5.$

So it never drops below $5$. A bound must also be attained to be the minimum: equality holds exactly when $3(x - 3)^2 = 0$, that is, at $x = 3$, and substituting confirms it:

$3(3 - 3)^2 + 5 = 0 + 5 = 5.$

The minimum is $5$, taken at $x = 3$ and nowhere else.

The path is the region between two rectangles, so its area is the outer rectangle's area minus the pool's.

The path adds width $x$ at each end of each dimension, so the outer rectangle measures $12 + 2x$ by $8 + 2x$. The pool itself is $12 \times 8 = 96$ square metres, so the path's area is

$(12 + 2x)(8 + 2x) - 96 = 100.$

Expand and collect:

$96 + 24x + 16x + 4x^2 - 96 = 100 \quad\Rightarrow\quad 4x^2 + 40x - 100 = 0.$

The leading coefficient is $4$ and the right side is $0$, so divide every term by $4$:

$x^2 + 10x - 25 = 0.$

Start from the monic equation and move the constant across:

$x^2 + 10x = 25.$

Half of $10$ is $5$, so the completing constant is $5^2 = 25$. Add it to BOTH sides:

$x^2 + 10x + 25 = 25 + 25 = 50 \quad\Rightarrow\quad (x + 5)^2 = 50.$

Take the square root of both sides, keeping both signs, and simplify the radical by pulling out the perfect square $25$:

$x + 5 = \pm\sqrt{50} = \pm 5\sqrt{2} \quad\Rightarrow\quad x = -5 \pm 5\sqrt{2}.$

Since $\sqrt{2} \approx 1.4142$, $5\sqrt{2} \approx 7.0711$, so the two solutions are

$x \approx -5 + 7.0711 = 2.07 \quad\text{and}\quad x \approx -5 - 7.0711 = -12.07,$

both in metres, to the nearest centimetre.

The equation was built from the geometry, but cannot remember that: it returns every number that satisfies it, and one of the two the situation has no room for.

A path's width is a length, so $x$ must be positive. The solution $x = -5 - 5\sqrt{2} \approx -12.07$ is negative, so it is discarded: a perfectly good root of the quadratic and a perfectly impossible path. The surviving solution is

$x = -5 + 5\sqrt{2} \approx 2.07 \text{ metres}.$

Check it against the budget rather than the algebra, so a modelling slip would still be caught. With $x = -5 + 5\sqrt{2}$, the outer rectangle measures $12 + 2x = 2 + 10\sqrt{2}$ by $8 + 2x = -2 + 10\sqrt{2}$ metres, so the paved region covers

$(2 + 10\sqrt{2})(-2 + 10\sqrt{2}) - 96 = (10\sqrt{2})^2 - 2^2 - 96 = 200 - 4 - 96 = 100 \text{ square metres},$

exactly the paving available. The width stands.

Test each line on its own terms.

Line 1 is sound: subtracting $5$ from both sides moves the constant across, treating both sides alike.

Line 2 is the first error. The student divides by $2$ to isolate $x^2$, but only the $2x^2$ actually gets divided; the $-8x$ and the $-5$ are left as they were, so line 2 is not equivalent to line 1. Dividing every term gives

$x^2 - 4x = -\frac{5}{2}.$

Line 3 is the second error. The completing constant is added to the left side only, which changes the equation. (The constant is also computed from the wrong $b$, but that follows from the first error: with the corrected line 2, half of $-4$ is $-2$ and the constant is $4$.) Added to both sides, the line should read

$x^2 - 4x + 4 = -\frac{5}{2} + 4.$

Divide EVERY term by the leading coefficient $2$:

$x^2 - 4x + \frac{5}{2} = 0 \quad\Rightarrow\quad x^2 - 4x = -\frac{5}{2}.$

Half of $-4$ is $-2$, so the completing constant is $(-2)^2 = 4$. Add it to both sides:

$x^2 - 4x + 4 = -\frac{5}{2} + 4 = \frac{3}{2} \quad\Rightarrow\quad (x - 2)^2 = \frac{3}{2}.$

The right side is POSITIVE, so the roots are real: the student's no-real-solutions verdict was an artefact of the two errors. Take the square root of both sides with a $\pm$, and simplify the root of the fraction over the perfect square $4$:

$x - 2 = \pm\sqrt{\frac{3}{2}} = \pm\sqrt{\frac{6}{4}} = \pm\frac{\sqrt{6}}{2}.$

Adding $2$ finishes it:

$x = 2 \pm \frac{\sqrt{6}}{2} = \frac{4 \pm \sqrt{6}}{2}.$

Numerically the roots are about $3.22$ and $0.78$; substituting either into $2x^2 - 8x + 5$ returns $0$.

An equation claims two quantities are equal. A step is legitimate exactly when it preserves that claim, so ask what it does to each side.

Add $16$ to the left side alone and the equation changes. If $x$ satisfied $x^2 - 8x = -5$, the left side was $-5$; after the addition it is $11$, not $-5$. The new equation $x^2 - 8x + 16 = -5$ demands something FALSE of the old solutions, and its own solutions are a different set with no obligation to overlap. The student's line 3 secretly moved the equation by $16$.

Adding the constant to both sides is different: it changes both quantities by the same amount, and equality survives that:

$\text{if } L = R, \text{ then } L + k = R + k.$

Every solution of the old equation is a solution of the new one, and since subtracting $k$ undoes the step, every solution of the new equation is a solution of the old. The two equations have exactly the same solution set, which licenses the move.

Move the constant across and complete the square. Half of $-6$ is $-3$, so the completing constant is $(-3)^2 = 9$, added to both sides:

$x^2 - 6x = -13 \quad\Rightarrow\quad x^2 - 6x + 9 = -13 + 9 = -4.$

The left side is a perfect square, and the number it landed on is NEGATIVE:

$(x - 3)^2 = -4.$

Take the square root of both sides with a $\pm$, converting the root of the negative with $i$: $\sqrt{-4} = i\sqrt{4} = 2i$.

$x - 3 = \pm 2i \quad\Rightarrow\quad x = 3 \pm 2i.$

The two solutions differ only in the sign of their imaginary parts.

Square the root first, using $i^2 = -1$ and the binomial-square expansion:

$(3 + 2i)^2 = 9 + 12i + 4i^2 = 9 + 12i - 4 = 5 + 12i.$

Substitute into the left-hand side and collect real and imaginary parts separately:

$(5 + 12i) - 6(3 + 2i) + 13 = 5 + 12i - 18 - 12i + 13 = 0 + 0i = 0.$

Both parts cancel, so $3 + 2i$ is a root.

The coefficient check runs on both roots at once. For a monic quadratic $x^2 + bx + c$, the roots sum to $-b$ and multiply to $c$; here $b = -6$ and $c = 13$, so they must sum to $6$ and multiply to $13$. The pair delivers exactly that, the imaginary parts cancelling in the sum and producing a real product through $i^2$:

$(3 + 2i) + (3 - 2i) = 6, \qquad (3 + 2i)(3 - 2i) = 9 - 4i^2 = 9 + 4 = 13.$

Both match: the product is a difference of squares in disguise, which is why the imaginary part disappears.

The completed form is an equivalent equation, not merely a convenient one, so anything true of the original is true of it:

$x^2 - 6x + 13 = 0 \quad\Longleftrightarrow\quad (x - 3)^2 = -4.$

Suppose some real $x$ satisfied it. Then $x - 3$ is real too, and a real square is never negative: whether $x - 3$ is positive, negative, or zero, its square is at least $0$.

So the left side would be at least $0$ while the right side is $-4$:

$(x - 3)^2 \ge 0 > -4$

which cannot hold. No real $x$ satisfies the equation, and the argument covers every real number, since the three sign cases exhaust them.

That is not a claim that the equation has no solutions: part A produced two. It is a claim about where they live, outside the reals, which is exactly what the $i$ in $3 \pm 2i$ records.

Do not test values of $c$. Complete the square once, with $c$ left as a letter, and every value is settled at once.

The linear coefficient is $-6$ whatever $c$ is, so the completing constant is $(-3)^2 = 9$, and adding it to both sides gives

$x^2 - 6x + 9 = 9 - c \quad\Longleftrightarrow\quad (x - 3)^2 = 9 - c.$

The roots' behaviour is now controlled by the sign of $9 - c$, the number whose square root is about to be taken. Three cases exhaust every real $c$.

If $9 - c > 0$, that is $c < 9$, the square root is a nonzero real and the $\pm$ produces two distinct REAL roots, $x = 3 \pm \sqrt{9 - c}$.

If $9 - c = 0$, that is $c = 9$, the root is $0$, so the two roots collapse into the single repeated real root $x = 3$.

If $9 - c < 0$, that is $c > 9$, the square root of a negative is $i$ times a positive real, and the $\pm$ produces the conjugate pair

$x = 3 \pm i\sqrt{c - 9},$

whose imaginary part $\sqrt{c - 9}$ is nonzero precisely because $c > 9$. So the roots form a conjugate pair in this case and no other: the claim is TRUE.

Check it against part A, where $c = 13$: since $13 > 9$, the claim predicts imaginary part $\sqrt{13 - 9} = 2$, matching the roots $3 \pm 2i$.