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

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Answer 1 · 2015-06-13T22:39:35

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

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

In general:

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

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

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

How do you complete the square for x2+7xx^2+7x?