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

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

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:

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

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 ($\scriptstyle x^{2}+2x\cdot 7$) corresponds to $\scriptstyle u^{2}+2uv$ in the elementary identity labeled (*) above. If $\scriptstyle x^{2}$ is $\scriptstyle u^{2}$ and $\scriptstyle 2x\cdot 7$ is $\scriptstyle 2uv$, then $\scriptstyle v$ must be 7. Therefore $\scriptstyle (u+v)^{2}$ must be $\scriptstyle (x+7)^{2}$. 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 $\scriptstyle 7^{2}$ inside the parentheses, and compensated (thus justifying the "=") by subtracting $\scriptstyle 3(7^{2})$ outside the parentheses. The expression inside the parentheses is now $\scriptstyle u^{2}+2uv+v^{2}$, 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

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

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

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