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

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

and completing the square is the way of filing in the blank between the brackets. Completing the square is used for solving quadratic equations (the proof of the well-known quadratic formula 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:

$3x^{2}+42x-5.\,$

We write

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

Now the expression (x² + 2x·7) corresponds to u² + 2uv in the elementary identity labeled (*) above. If x² is u² and 2x·7 is 2uv, then v must be 7. Therefore (u + v)² must be (x + 7). So we continue:

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

Now we have added 7² inside the parentheses, and compensated (thus justifying the "=") by subtracting 3(7²) outside the parentheses. The expression inside the parentheses is now u² + 2uv + v², and by the elementary identity labeled (*) above, it is therefore equal to (u + v)², i.e. to (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 abstracly

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 $\Delta <math>standsforthewell-known''discriminant''ofthepolynomial,thatis<math>\Delta =b^{2}-4ac.$

Indeed, we have

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

The last expression inside parentheses above corresponds to u² + 2uv in the identity labelled (*) above. We need to add the third term, v²:

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