Solution 2.3:2c
From Förberedande kurs i matematik 1
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| - | {{ | + | We start by completing the square of the left-hand side, |
| - | < | + | |
| - | {{ | + | {{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. | ||
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.
