How to "Re-write completing the square": $x^2+x+1$

Solution 1:

Remember the formula for the square of a binomial: $$(a+b)^2 = a^2 + 2ab + b^2.$$

Now, when you see $x^2+x+1$, you want to think of $x^2+x$ as the first two terms you get in expanding the binomial $(x+c)^2$ for some $c$; that is, $$x^2 + x + \cdots = (x+c)^2.$$

Since the middle term should be $2cx$, and you have $x$, that means that you want $2c=1$, or $c=\frac{1}{2}$.

But if you have $(x+\frac{1}{2})^2$, you get $x^2 + x + \frac{1}{4}$. Since all you have is $x^2+x$, you complete the square by adding the missing "$\frac{1}{4}$". Since you are not allowed to just add constants willy-nilly, you must also cancel it out by subtracting $\frac{1}{4}$. So: \begin{align*} x^2 + x + 2 &= (x^2 + x + \cdots) + 1\\ &=\left( x^2 + 2\left(\frac{1}{2}\right)x + \cdots \right) + 1 &&\mbox{figuring out what $c$ is}\\ &= \left(x^2 + 2\left(\frac{1}{2}\right)x + \left(\frac{1}{2}\right)^2\right) -\frac{1}{4} + 1 &&\mbox{completing the square}\\ &= \left(x+\frac{1}{2}\right)^2 + \frac{3}{4}. \end{align*}

Solution 2:

"Completing the square" is a standard step in solving a quadratic equation. To see how it helps, consider the following: The general formulation of a quadratic equation is $ax^2+bx+c = 0$ with $a \neq 0$. Let us say we are tasked with solving this equation, i.e., finding values of $x$ that satisfy this equation. To start with, note that this equation is easy to solve if $b=0$. Then the equation looks like $ax^2 + c = 0$ which would simplify to $x = \pm \sqrt{\frac{-c}{a}}$.

"Completing the square" is a step that takes a general quadratic and reduces it to the form of the simple quadratic above. It does this with a substitution $y = x + \frac{b}{2a}$. Then, $ay^2 = ax^2 + bx + \frac{b^2}{4a}$ which means the general quadratic can be written as $ay^2-\frac{b^2}{4a} + c = 0$ or equivalently as $$ a(x+\frac{b}{2a})^2 = \frac{b^2}{4a} - c $$

This equation is easy to solve and yields the two roots of the general quadratic equation. Note that the above equation is equivalent to the one we started with ($ax^2+bx+c = 0$) for the purposes of finding the roots. This process of rewriting is called completing the square. This is the point behind rewriting $x^2+x+1$ as $(x+\frac{1}{2})^2 + \frac{3}{4}$.

Solution 3:

One can rewrite a degree $\rm\:n > 1\:$ polynomial $\rm\ f(x)\ =\ x^n + b\ x^{n-1} +\ \cdots\ $ into a form such that its two highest degree terms are "absorbed" into a perfect $\rm\: n$'th power of a linear polynomial, namely

$\rm\quad\quad f(x)\ =\ (x + b/n)^n\ -\ g(x)\ \ $ where $\rm\ \ g(x)\ =\ (x+b/n)^n - f(x)\ $ has degree $\rm\:\le\: n-2$

When $\rm\ n = 2\ $ this is called completing the square - esp. when used to solve a quadratic equation.$\ \ $ If $\rm\ \ g(x)\ = g\ $ is constant (as is always true when $\rm\ n = 2\:$) then this yields a closed form for the roots of $\rm\:\ f(x)\:,\ $ namely $\rm\ x\ =\ \sqrt[n]{g}-b/2\:.$