Solution 2.3:6c - Revision history

Tek at 11:45, 29 September 2008

Tek — Mon, 29 Sep 2008 11:45:27 GMT

←Older revision		Revision as of 11:45, 29 September 2008
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	If we complete the square of the expression, we have that		If we complete the square of the expression, we have that

		+	{{Displayed math\|\|<math>\begin{align}
		+	x^{2} - 5x + 7 &= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \Bigl(\frac{5}{2}\Bigr)^{2} + 7\\[5pt]
		+	&= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \frac{25}{4} + \frac{28}{4}\\[5pt]
		+	&= \Bigl(x-\frac{5}{2}\Bigr)^{2} + \frac{3}{4}
		+	\end{align}</math>}}

-	~~<math>\begin{align}~~	+	and because <math>\bigl(x-\tfrac{5}{2}\bigr)^{2}</math> is a quadratic, this term assumes a minimal value zero when <math>x=5/2\,</math>. This shows that the polynomial's smallest value is <math>\tfrac{3}{4}</math>.
-	~~& x^{2}-5x+7=\left( x-\frac{5}{2} \right)^{2}-\left( \frac{5}{2} \right)^{2}+7 \\~~	+
-	~~& =\left( x-\frac{5}{2} \right)^{2}-\frac{25}{4}+\frac{28}{4}=\left( x-\frac{5}{2} \right)^{2}+\frac{3}{4} \\~~	+
-	~~\end{align}</math>~~	+
-		+
-		+
-	and because	+
-	<math>\~~left~~( x-\~~frac~~{5}{2} \~~right~~)^{2}</math>	+
-	is a quadratic, this term ~~is at least equal to~~ zero when	+
-	<math>x={5}/{2}\;</math>. This shows that the polynomial's smallest value is	+
-	<math>\~~frac~~{3}{4}</math>.	+

Ian — Sun, 21 Sep 2008 11:03:37 GMT

←Older revision		Revision as of 11:03, 21 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	If we complete the square of the expression, we have that
-	<~~center~~> ~~[[Image:2_3_6c.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	<math>\begin{align}
		+	& x^{2}-5x+7=\left( x-\frac{5}{2} \right)^{2}-\left( \frac{5}{2} \right)^{2}+7 \\
		+	& =\left( x-\frac{5}{2} \right)^{2}-\frac{25}{4}+\frac{28}{4}=\left( x-\frac{5}{2} \right)^{2}+\frac{3}{4} \\
		+	\end{align}</math>
		+
		+
		+	and because
		+	<math>\left( x-\frac{5}{2} \right)^{2}</math>
		+	is a quadratic, this term is at least equal to zero when
		+	<math>x={5}/{2}\;</math>. This shows that the polynomial's smallest value is
		+	<math>\frac{3}{4}</math>.

Tekbot — Tue, 09 Sep 2008 09:02:43 GMT

←Older revision

Revision as of 09:02, 9 September 2008

Tekbot — Thu, 21 Aug 2008 06:48:29 GMT

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Oskar — Tue, 01 Apr 2008 10:14:00 GMT

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