# How do you complete the square for x^2+20x?

May 29, 2015

Since the coefficient of ${x}^{2}$ is 1, we can take the coefficient of $x$, the 20, divide it by 2 to get 10, and then square that number to get 100. Now add zero by adding and subtracting 100 to get:

${x}^{2} + 20 x = \left({x}^{2} + 20 x + 100\right) - 100$

But the expression ${x}^{2} + 20 x + 100$ is a perfect square:
${x}^{2} + 20 x + 100 = {\left(x + 10\right)}^{2}$.

Hence,

${x}^{2} + 20 x = \left({x}^{2} + 20 x + 100\right) - 100 = {\left(x + 10\right)}^{2} - 100$

One purpose of this procedure is to help you graph the function $f \left(x\right) = {x}^{2} + 20 x$. Since $f \left(x\right) = {\left(x + 10\right)}^{2} - 100$ as well, you can see that the minimum value of $f \left(x\right)$ is $- 100$ when $x = - 10$ (note that the coefficient of ${\left(x + 10\right)}^{2}$ is 1, which is positive, so the vertex is a minimum rather than a maximum). The vertex of the parabola that is the graph of $f$ is at $\left(x , y\right) = \left(- 10 , - 100\right)$. The graph opens upward, and you can find other things to help you graph it, such as $x$ and $y$-intercepts (the $y$-intercept is $f \left(0\right) = 0$ and the $x$-intercepts are $x = 0$ and $x = - 20$ since $f \left(x\right) = {x}^{2} + 20 x = x \left(x + 20\right)$.

Can you use the method above to complete the square in the following general case? $f \left(x\right) = {x}^{2} + b x + c$? Give it a try and see if you can figure out where the vertex of the graph of $f$ is after doing so (its coordinates depend on $b$ and $c$).