Processing Math: Done

From Förberedande kurs i matematik 1

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-	{{NAVCONTENT_START}}	+	We apply the standard formula for completing the square,
-	<center> [[Image:2_3_1d.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>x^{2}+ax = \Bigl(x+\frac{a}{2}\Bigr)^{2} - \Bigl(\frac{a}{2}\Bigr)^{2}\,\textrm{,}</math>}}
		+
		+	on our expression and this gives
		+
		+	{{Displayed math||<math>x^{2}+5x = \Bigl(x+\frac{5}{2}\Bigr)^{2} - \Bigl(\frac{5}{2}\Bigr)^{2} = \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{25}{4}\,\textrm{.}</math>}}
		+
		+	The whole expression becomes
		+
		+	{{Displayed math||<math>\begin{align}
		+	x^{2}+5x+3
		+	&= \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{25}{4}+3\\[5pt]
		+	&= \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{25}{4} + \frac{12}{4}\\[5pt]
		+	&= \Bigl(x+\frac{5}{2}\Bigr)^{2} + \frac{12-25}{4}\\[5pt]
		+	&= \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{13}{4}\,\textrm{.}
		+	\end{align}</math>}}
		+
		+	A quick check shows that we have calculated correctly
		+
		+	{{Displayed math||<math>\begin{align}
		+	\Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{13}{4}
		+	&= x^{2} + 2\cdot\frac{5}{2}\cdot x + \Bigl(\frac{5}{2}\Bigr)^{2} - \frac{13}{4}\\[5pt]
		+	&= x^{2} + 5x + \frac{25}{4} - \frac{13}{4}\\[5pt]
		+	&= x^{2} + 5x + \frac{12}{4}\\[5pt]
		+	&= x^{2}+5x+3\,\textrm{.}
		+	\end{align}</math>}}

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Current revision

We apply the standard formula for completing the square,

x2+ax=x+2a2−2a2,

on our expression and this gives

x2+5x=x+252−252=x+252−425.

The whole expression becomes

x2+5x+3=x+252−425+3=x+252−425+412=x+252+412−25=x+252−413.

A quick check shows that we have calculated correctly

x+252−413=x2+225x+252−413=x2+5x+425−413=x2+5x+412=x2+5x+3.