How do you complete the square for $x^{2} + 11 x$ $x^2 + 11x$ ?

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How do you complete the square for $x^{2} + 11 x$ $x^2 + 11x$ ?

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Answer 1 · 2015-07-19T09:32:29

$x^{2} + 11 x = {(x + \frac{11}{2})}^{2} - \frac{11^{2}}{2^{2}} = {(x + \frac{11}{2})}^{2} - \frac{121}{4}$ $x^2+11x = (x+11/2)^2-11^2/(2^2) = (x+11/2)^2-121/4$

Explanation:

In general,

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

Notice the $\frac{b}{2 a}$ $b/(2a)$ term.

In our case $a = 1$ $a=1$ and $b = 11$ $b=11$ hence the $\frac{11}{2}$ $11/2$

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How do you complete the square for $x^{2} + 11 x$ $x^2 + 11x$ ?

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