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

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Answer 1 · 2015-06-13T22:39:35

Note that $(x+7/2)^2 = x^2+7x+49/4$

So $x^2 + 7x = (x+7/2)^2 - 49/4$

In general:

$ax^2+bx+c = a(x+b/(2a))^2 + (c - b^2/(4a))$

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

$x^2+bx+c = (x+b/2)^2 + (c - b^2/4)$

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