Completing the Square

Summary: Completing the square transforms a mixt quadratic $x^2+px=q$ into a perfect-square equation by adding $\frac{1}{4}{p}^2$ to both sides, yielding the quadratic formula and forming the basis for all of Euler’s quadratic theory.

Sources: chapter-1.4.6

Last updated: 2026-05-03

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The technique

Given the equation $x^2+px=q$ (source: chapter-1.4.6, §640–641):

1. Observe that $(x+\frac{1}{2}p{)}^2=x^2+px+\frac{1}{4}{p}^2$.
2. Add $\frac{1}{4}{p}^2$ to both sides:

$x^2+px+\frac{1}{4}{p}^2=q+\frac{1}{4}{p}^2$

3. The left side is now a perfect square:

$(x+\frac{1}{2}p{)}^2=\frac{1}{4}{p}^2+q$

4. Extract the square root (both signs):

$x+\frac{1}{2}p=\pm \sqrt{\frac{1}{4}{p}^2+q}$

5. Solve for $x$:

$x=-\frac{1}{2}p\pm \sqrt{\frac{1}{4}{p}^2+q}$

Geometric intuition

Adding $\frac{1}{4}{p}^2$ “completes” the square $(x+\frac{p}{2}{)}^2$. The name comes from the geometric image of completing a square of side $x+p/2$ from a rectangle of area $px$ and a square of area $x^2$.

Equivalent substitution proof (§645)

Let $x=y+\frac{1}{2}p$. Substituting into $x^2=px+q$:

$y^2+py+\frac{1}{4}{p}^2=py+\frac{1}{2}{p}^2+q$

Cancelling $py$ and subtracting $\frac{1}{4}{p}^2$:

$y^2=\frac{1}{4}{p}^2+q$

which is a pure quadratic in $y$. Since $x=y+\frac{1}{2}p$, the same formula results.

The standard quadratic formula

For $a{x}^2+bx+c=0$ (divide by $a$ first, then apply the technique):

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Applications

• All quadratic equations (Chapters 1.4.5–1.4.6)
• Extracting roots of polygonal numbers (Chapter 1.4.7) — each polygonal class yields a quadratic completed in the same way
• Extracting square-roots-of-binomials (Chapter 1.4.8) — the system $x+y=a$, $xy=b/4$ is resolved by completing the square on $x-y$

Related pages

• quadratic-equations
• ch1.4.6-mixt-quadratic-equations
• ch1.4.5-pure-quadratic-equations
• ch1.4.7-extraction-roots-polygonal-numbers
• discriminant
• vieta-formulas