From Förberedande kurs i matematik 2

Revision as of 13:39, 7 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:3_3_3b.gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (13:33, 30 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	When we complete the square, we replace all <math>z</math>-terms in the second-degree expression with a quadratic term which contains <math>z</math>, according to the formula
-	<center> [[Bild:3_3_3b.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>z^2+az = \Bigl(z+\frac{a}{2}\Bigr)^2 - \Bigl(\frac{a}{2}\Bigr)^2\,\textrm{.}</math>}}
		+
		+	In our case, we set <math>a=3i</math> in order to complete the square,
		+
		+	{{Displayed math||<math>\begin{align}
		+	z^2+3iz-\frac{1}{4}
		+	&= \Bigl(z+\frac{3}{2}\,i\Bigr)^2 - \Bigl(\frac{3}{2}\,i\Bigr)^2 - \frac{1}{4}\\[5pt]
		+	&= \Bigl(z+\frac{3}{2}\,i\Bigr)^2 - \frac{9}{4}\cdot (-1) - \frac{1}{4}\\[5pt]
		+	&= \Bigl(z+\frac{3}{2}\,i\Bigr)^2+2\,\textrm{.}
		+	\end{align}</math>}}

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Current revision

When we complete the square, we replace all \displaystyle z-terms in the second-degree expression with a quadratic term which contains \displaystyle z, according to the formula

\displaystyle z^2+az = \Bigl(z+\frac{a}{2}\Bigr)^2 - \Bigl(\frac{a}{2}\Bigr)^2\,\textrm{.}

In our case, we set \displaystyle a=3i in order to complete the square,

\displaystyle \begin{align} z^2+3iz-\frac{1}{4} &= \Bigl(z+\frac{3}{2}\,i\Bigr)^2 - \Bigl(\frac{3}{2}\,i\Bigr)^2 - \frac{1}{4}\\[5pt] &= \Bigl(z+\frac{3}{2}\,i\Bigr)^2 - \frac{9}{4}\cdot (-1) - \frac{1}{4}\\[5pt] &= \Bigl(z+\frac{3}{2}\,i\Bigr)^2+2\,\textrm{.} \end{align}