How do you complete the square for $x^{2} + 8 x$ $x^2+8x$ ?

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How do you complete the square for $x^{2} + 8 x$ $x^2+8x$ ?

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Answer 1 · 2018-05-15T21:31:50

$x^{2} + 8 x + 16$ $x^2+8x+16$

Explanation:

$since the coefficient of the x^{2} term is 1$ $"since the coefficient of the "x^2" term is 1"$

$add {(\frac{1}{2} coefficient of the x-term)}^{2} to$ $"add "(1/2"coefficient of the x-term")^2" to"$
$x^{2} + 8 x$ $x^2+8x$

$\Rightarrow x^{2} + 8 x {+ 4}^{2} = x^{2} + 8 x + 16 = {(x + 4)}^{2}$ $rArrx^2+8xcolor(red)(+4)^2=x^2+8x+16=(x+4)^2$

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How do you complete the square for $x^{2} + 8 x$ $x^2+8x$ ?

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How do you complete the square for $x^{2} + 8 x$ $x^2+8x$ ?