Solution 3.3:3a
From Förberedande kurs i matematik 2
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- | {{ | + | When we complete the square of the second degree expression, <math>z^2+2z+3</math>, we start with the forumla |
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- | {{ | + | {{Displayed math||<math>(z+a)^2 = z^2+2az+a^2</math>}} |
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+ | which we write as | ||
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+ | {{Displayed math||<math>(z+a)^2-a^2 = z^2+2az</math>}} | ||
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+ | and adapt the constant to be <math>a=1</math> so that the terms <math>z^2+2z</math> are equal to <math>z^2+2az</math>, and therefore can be written as <math>(z+1)^2-1^2</math>. The whole calculation becomes | ||
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+ | {{Displayed math||<math>\underline{z^2+2z\vphantom{()}}+3 = \underline{(z+1)^2-1^2}+3 = (z+1)^2 + 2\,\textrm{.}</math>}} | ||
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+ | The underlined terms show the actual completion of the square. |
Current revision
When we complete the square of the second degree expression, \displaystyle z^2+2z+3, we start with the forumla
\displaystyle (z+a)^2 = z^2+2az+a^2 |
which we write as
\displaystyle (z+a)^2-a^2 = z^2+2az |
and adapt the constant to be \displaystyle a=1 so that the terms \displaystyle z^2+2z are equal to \displaystyle z^2+2az, and therefore can be written as \displaystyle (z+1)^2-1^2. The whole calculation becomes
\displaystyle \underline{z^2+2z\vphantom{()}}+3 = \underline{(z+1)^2-1^2}+3 = (z+1)^2 + 2\,\textrm{.} |
The underlined terms show the actual completion of the square.