From Förberedande kurs i matematik 1

Revision as of 06:56, 21 August 2008 (edit) Tekbot (Talk | contribs) m (Robot: Automated text replacement (-[[Bild: +[[Image:)) ← Previous diff		Current revision (12:42, 1 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	Square both sides of the equation so that the root sign disappears,
-	<center> [[Image:3_2_4.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>1-x = (2-x)^2\quad \Leftrightarrow \quad 1-x = 4-4x+x^2</math>}}
		+
		+	and then solve the resulting second-order equation by completing the square,
		+
		+	{{Displayed math||<math>\begin{align}
		+	x^{2}-3x+3 &= 0\,,\\[5pt]
		+	\Bigl(x-\frac{3}{2}\Bigr)^{2} - \Bigl(\frac{3}{2}\Bigr)^{2} + 3 &= 0\,,\\[5pt]
		+	\Bigl(x-\frac{3}{2}\Bigr)^{2} - \frac{9}{4} + \frac{12}{4} &= 0\,,\\[5pt]
		+	\Bigl(x-\frac{3}{2}\Bigr)^{2} + \frac{3}{4} &= 0\,\textrm{.}
		+	\end{align}</math>}}
		+
		+	As can be seen, the second-order equation does not have any solutions (the left-hand side is always greater than or equal to 3/4, regardless of how ''x'' is chosen) so, the original root equation does not have any solutions.

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Current revision

Square both sides of the equation so that the root sign disappears,

\displaystyle 1-x = (2-x)^2\quad \Leftrightarrow \quad 1-x = 4-4x+x^2

and then solve the resulting second-order equation by completing the square,

\displaystyle \begin{align} x^{2}-3x+3 &= 0\,,\\[5pt] \Bigl(x-\frac{3}{2}\Bigr)^{2} - \Bigl(\frac{3}{2}\Bigr)^{2} + 3 &= 0\,,\\[5pt] \Bigl(x-\frac{3}{2}\Bigr)^{2} - \frac{9}{4} + \frac{12}{4} &= 0\,,\\[5pt] \Bigl(x-\frac{3}{2}\Bigr)^{2} + \frac{3}{4} &= 0\,\textrm{.} \end{align}

As can be seen, the second-order equation does not have any solutions (the left-hand side is always greater than or equal to 3/4, regardless of how x is chosen) so, the original root equation does not have any solutions.