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Solution 2.3:2c

From Förberedande kurs i matematik 1

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m (Lösning 2.3:2c moved to Solution 2.3:2c: Robot: moved page)
Current revision (07:38, 29 September 2008) (edit) (undo)
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We start by completing the square of the left-hand side,
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<center> [[Image:2_3_2c.gif]] </center>
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{{NAVCONTENT_STOP}}
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{{Displayed math||<math>\begin{align}
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y^{2}+3y+4 &= \Bigl(y+\frac{3}{2}\Bigr)^{2} - \Bigl(\frac{3}{2}\Bigr)^{2}+4\\[5pt]
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&= \Bigl(y+\frac{3}{2}\Bigr)^{2} - \frac{9}{4} + \frac{16}{4}\\[5pt]
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&= \Bigl(y+\frac{3}{2}\Bigr)^{2} + \frac{7}{4}\,\textrm{.}
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\end{align}</math>}}
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The equation is then
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{{Displayed math||<math>\left( y+\frac{3}{2} \right)^{2}+\frac{7}{4}=0\,\textrm{.}</math>}}
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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.

Current revision

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

y2+3y+4=y+232232+4=y+23249+416=y+232+47.

The equation is then

y+232+47=0. 

The first term y+232  is always greater than or equal to zero because it is a square and 47 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.