How do you complete the square for x^2+7x?

1 Answer
Jun 13, 2015

Note that ${\left(x + \frac{7}{2}\right)}^{2} = {x}^{2} + 7 x + \frac{49}{4}$

So ${x}^{2} + 7 x = {\left(x + \frac{7}{2}\right)}^{2} - \frac{49}{4}$

Explanation:

In general:

$a {x}^{2} + b x + c = a {\left(x + \frac{b}{2 a}\right)}^{2} + \left(c - {b}^{2} / \left(4 a\right)\right)$

In the common cases where $a = 1$ this simplifies to:

${x}^{2} + b x + c = {\left(x + \frac{b}{2}\right)}^{2} + \left(c - {b}^{2} / 4\right)$