Completing the Square is the habit of rewriting quadratics into square-plus-constant form to reveal minima, distances, or constraints. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.
Any quadratic can be rewritten as a perfect square plus a constant, which instantly exposes its vertex, minimum or maximum, and how to solve it.
Why It Works
1
Factor $a$ out of the variable terms: $ax^2 + bx + c = a\left(x^2 + \frac{b}{a}x\right) + c$.
2
Inside the parentheses, add and subtract the square of half the linear coefficient: $x^2 + \frac{b}{a}x = \left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2}$.
Since $-(x - 4)^2 \le 0$, the largest value is $11$, attained at $x = 4$.
Answer:
Maximum value $11$, at $x = 4$.
Contest level
Solve $x^2 + 6x + 7 = 0$ by completing the square.
1
Half of $6$ is $3$, and $3^2 = 9$: write $x^2 + 6x + 7 = (x + 3)^2 - 9 + 7 = (x + 3)^2 - 2$.
2
Set equal to zero: $(x + 3)^2 = 2$.
3
Take square roots: $x + 3 = \pm\sqrt 2$, so $x = -3 \pm \sqrt 2$.
Answer:
$x = -3 + \sqrt 2$ or $x = -3 - \sqrt 2$.
Olympiad / Challenge
Find all real pairs $(x, y)$ satisfying $x^2 + y^2 + 1 = xy + x + y$.
1
Multiply both sides by $2$ to clear fractions when completing squares: $2x^2 + 2y^2 + 2 = 2xy + 2x + 2y$, i.e. $2x^2 + 2y^2 - 2xy - 2x - 2y + 2 = 0$.
2
Regroup as a sum of three squares: $(x - y)^2 + (x - 1)^2 + (y - 1)^2 = 0$. (Expand to confirm: $x^2 - 2xy + y^2$, $x^2 - 2x + 1$, $y^2 - 2y + 1$ sum to exactly the left side.)
3
A sum of real squares is $0$ only if each is $0$: $x = y$, $x = 1$, $y = 1$, forcing $x = y = 1$.
Answer:
$(x, y) = (1, 1)$ is the only real solution.
Going Deeper
Generalization: completing the square diagonalizes any quadratic form. In two variables $x^2 + y^2 + z^2 \pm \cdots$ it expresses an expression as a sum of squares (SOS), the standard route to proving nonnegativity and locating equality.
Where it appears: it derives the quadratic formula itself, finds parabola vertices on the AMC, converts conic equations to standard form, and (the SOS trick above) cracks olympiad problems where a single equation pins down several variables.
Pitfall: when $a \ne 1$ you must factor $a$ out of the variable terms FIRST — completing the square on $2x^2 + bx$ as if the leading coefficient were $1$ is wrong. And the constant added inside the parentheses gets multiplied by $a$ when you redistribute, a frequent sign/scaling slip.
Spot the Signal
Look for problems where the key step is rewriting quadratics into square-plus-constant form to reveal minima, distances, or constraints.
You can describe the hard part as rewriting quadratics into square-plus-constant form to reveal minima, distances, or constraints, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.
Learn the Move
Start by identify the expression or equation that calls for completing the square, then rewrite the givens around it.
Name the relation that makes Completing the Square legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.
Avoid These Traps
Do not use Completing the Square just because the surface looks familiar; verify the required condition first.
Applying Completing the Square because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.
MathGrit Coach Note
Let completing the square reveal the algebraic structure; then compute only what remains.
Try it on:
Practice a contest problem where the key step is rewriting quadratics into square-plus-constant form to reveal minima, distances, or constraints.