Completing the square

In algebra, completing the square is a way of rewriting a quadratic polynomial as the sum of a constant and a constant multiple of the square of a first-degree polynomial. Thus one has


 * $$ ax^2 + bx + c = a(x+\cdots)^2 + \text{constant} $$

and completing the square is the way of filling in the blank between the brackets. Completing the square is used for solving quadratic equations (the proof of the well-known formula for the general solution consists of completing the square). The technique is also used to find the maximum or minimum value of a quadratic function, or in other words, the vertex of a parabola.

The technique relies on the elementary algebraic identity


 * $$ (u + v)^2 = u^2 + 2uv + v^2.\qquad\qquad(*) $$

Concrete examples
We want to fill in this blank:


 * $$ 3x^2 + 42x - 5 = 3(x+\cdots)^2 + \text{constant}. $$

We write



\begin{align} 3x^2 + 42x - 5 & {} = 3(x^2 + 14x) - 5 \\ & {} = 3(x^2 + 2x\cdot 7) - 5. \end{align} $$

Now the expression ($$ \scriptstyle x^2 + 2 x \cdot 7$$) corresponds to $$ \scriptstyle u^2 + 2 u v$$ in the elementary identity labeled (*) above. If $$ \scriptstyle x^2 $$ is $$ \scriptstyle u^2 $$ and $$ \scriptstyle 2 x \cdot 7 $$ is $$ \scriptstyle 2 u v $$, then $$ \scriptstyle v $$ must be 7. Therefore $$ \scriptstyle ( u + v )^2 $$ must be $$ \scriptstyle ( x + 7 )^2 $$. So we continue:



\begin{align} & {} 3(x^2 + 2x\cdot 7) - 5 \\ & {} = 3\left(x^2 + 2x\cdot 7 + 7^2\right) - 5 - 3(7^2), \end{align} $$

Now we have added $$ \scriptstyle 7^2 $$ inside the parentheses, and compensated (thus justifying the "=") by subtracting $$ \scriptstyle 3 (7^2) $$ outside the parentheses. The expression inside the parentheses is now $$ \scriptstyle u^2 + 2 u v + v^2 $$, and by the elementary identity labeled (*) above, it is therefore equal to $$ \scriptstyle ( u + v )^2 $$, i.e. to $$ \scriptstyle ( x + 7 )^2 $$. So now we have


 * $$ 3(x + 7)^2 - 5 - 3(7^2) = 3(x + 7)^2 - 152.\, $$

Thus we have the equality


 * $$ 3x^2 + 42x - 5 = 3(x + 7)^2 - 152.\, $$

More abstractly
It is possible to give a closed formula for the completion in terms of the coefficients a, b and c. Namely,
 * $$ ax^2+bx + c = a\left(x + \frac{b}{2a}\right)^2 - \frac{\Delta}{4a}, $$

where $$ \scriptstyle \Delta $$ stands for the well-known discriminant of the polynomial, that is $$ \scriptstyle \Delta = b^2 - 4ac $$.

Indeed, we have



\begin{align} ax^2 + bx + c & {} = a\left( x^2 + \frac{b}{a}x \right) + c \\ & {} = a\left(x^2 + 2\frac{b}{2a}x\right) + c. \end{align} $$

The last expression inside parentheses above corresponds to $$ \scriptstyle u^2 + 2 u v $$ in the identity labelled (*) above. We need to add the third term, $$ \scriptstyle v^2 $$:



\begin{align} ax^2 + bx + c & {} = a\left( x^2 + \frac{b}{a}x \right) + c \\ & {} = a\left(x^2 + 2\frac{b}{2a}x +\left(\frac{b}{2a}\right)^2 \right) + c - a\left(\frac{b}{2a}\right)^2 \\ & {} = a\left(x + \frac{b}{2a}\right)^2 + c - \frac{b^2}{4a} \\ & {} = a\left(x + \frac{b}{2a}\right)^2 + \frac{4ac - b^2}{4a}. \end{align} $$