From Förberedande kurs i matematik 1

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-	{{NAVCONTENT_START}}	+	We solve the second order equation by combining together the ''x''²- and ''x''-terms by completing the square to obtain a quadratic term, and then solve the resulting equation by taking the root.
-	<center> [[Bild:2_3_2a.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	By completing the square, the left-hand side becomes
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		+	{{Displayed math||<math>\underline{x^{2}-4x\vphantom{()}}+3 = \underline{(x-2)^{2}-2^{2}}+3 = (x-2)^{2}-1\,\textrm{,}</math>}}
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		+	where the underlined part on the right-hand side is the actual completed square. The equation can therefore be written as
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		+	{{Displayed math||<math>(x-2)^{2}-1 = 0</math>}}
		+
		+	which we solve by moving the "1" on the right-hand side and taking the square root. This gives the solutions:
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		+	:*<math>x-2=\sqrt{1}=1\,,\ </math> i.e. <math>x=2+1=3\,,</math>
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		+	:*<math>x-2=-\sqrt{1}=-1\,,\ </math> i.e. <math>x=2-1=1\,\textrm{.}</math>
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		+	Because it is easy to make a mistake, we check the answer by substituting <math>x=1</math> and <math>x=3</math> into the original equation:
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		+	:*''x''&nbsp;=&nbsp;1: <math>\ \text{LHS} = 1^{2}-4\cdot 1+3 = 1-4+3 = 0 = \text{RHS,}</math>
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		+	:*''x''&nbsp;=&nbsp;3: <math>\ \text{LHS} = 3^{2}-4\cdot 3+3 = 9-12+3 = 0 = \text{RHS.}</math>

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Current revision

We solve the second order equation by combining together the x²- and x-terms by completing the square to obtain a quadratic term, and then solve the resulting equation by taking the root.

By completing the square, the left-hand side becomes

\displaystyle \underline{x^{2}-4x\vphantom{()}}+3 = \underline{(x-2)^{2}-2^{2}}+3 = (x-2)^{2}-1\,\textrm{,}

where the underlined part on the right-hand side is the actual completed square. The equation can therefore be written as

\displaystyle (x-2)^{2}-1 = 0

which we solve by moving the "1" on the right-hand side and taking the square root. This gives the solutions:

• \displaystyle x-2=\sqrt{1}=1\,,\ i.e. \displaystyle x=2+1=3\,,

• \displaystyle x-2=-\sqrt{1}=-1\,,\ i.e. \displaystyle x=2-1=1\,\textrm{.}

Because it is easy to make a mistake, we check the answer by substituting \displaystyle x=1 and \displaystyle x=3 into the original equation:

• x = 1: \displaystyle \ \text{LHS} = 1^{2}-4\cdot 1+3 = 1-4+3 = 0 = \text{RHS,}

• x = 3: \displaystyle \ \text{LHS} = 3^{2}-4\cdot 3+3 = 9-12+3 = 0 = \text{RHS.}