From Förberedande kurs i matematik 1

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-	{{NAVCONTENT_START}}	+	We start by completing the square of the left-hand side,
-	<center> [[Image:2_3_2c.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\begin{align}
		+	y^{2}+3y+4 &= \Bigl(y+\frac{3}{2}\Bigr)^{2} - \Bigl(\frac{3}{2}\Bigr)^{2}+4\\[5pt]
		+	&= \Bigl(y+\frac{3}{2}\Bigr)^{2} - \frac{9}{4} + \frac{16}{4}\\[5pt]
		+	&= \Bigl(y+\frac{3}{2}\Bigr)^{2} + \frac{7}{4}\,\textrm{.}
		+	\end{align}</math>}}
		+
		+	The equation is then
		+
		+	{{Displayed math||<math>\left( y+\frac{3}{2} \right)^{2}+\frac{7}{4}=0\,\textrm{.}</math>}}
		+
		+	The first term <math>\bigl(y+\tfrac{3}{2}\bigr)^{2}</math> is always greater than or equal to zero because it is a square and <math>\tfrac{7}{4}</math> is a positive number. This means that the left hand side cannot be zero, regardless of how ''y'' is chosen. The equation has no solution.

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Current revision

We start by completing the square of the left-hand side,

\displaystyle \begin{align} y^{2}+3y+4 &= \Bigl(y+\frac{3}{2}\Bigr)^{2} - \Bigl(\frac{3}{2}\Bigr)^{2}+4\\[5pt] &= \Bigl(y+\frac{3}{2}\Bigr)^{2} - \frac{9}{4} + \frac{16}{4}\\[5pt] &= \Bigl(y+\frac{3}{2}\Bigr)^{2} + \frac{7}{4}\,\textrm{.} \end{align}

The equation is then

\displaystyle \left( y+\frac{3}{2} \right)^{2}+\frac{7}{4}=0\,\textrm{.}

The first term \displaystyle \bigl(y+\tfrac{3}{2}\bigr)^{2} is always greater than or equal to zero because it is a square and \displaystyle \tfrac{7}{4} is a positive number. This means that the left hand side cannot be zero, regardless of how y is chosen. The equation has no solution.