Solution 2.3:1d
From Förberedande kurs i matematik 1
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| - | {{ | + | We apply the standard formula for completing the square, |
| - | < | + | |
| - | {{ | + | {{Displayed math||<math>x^{2}+ax = \Bigl(x+\frac{a}{2}\Bigr)^{2} - \Bigl(\frac{a}{2}\Bigr)^{2}\,\textrm{,}</math>}} |
| + | |||
| + | on our expression and this gives | ||
| + | |||
| + | {{Displayed math||<math>x^{2}+5x = \Bigl(x+\frac{5}{2}\Bigr)^{2} - \Bigl(\frac{5}{2}\Bigr)^{2} = \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{25}{4}\,\textrm{.}</math>}} | ||
| + | |||
| + | The whole expression becomes | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | x^{2}+5x+3 | ||
| + | &= \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{25}{4}+3\\[5pt] | ||
| + | &= \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{25}{4} + \frac{12}{4}\\[5pt] | ||
| + | &= \Bigl(x+\frac{5}{2}\Bigr)^{2} + \frac{12-25}{4}\\[5pt] | ||
| + | &= \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{13}{4}\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | A quick check shows that we have calculated correctly | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{13}{4} | ||
| + | &= x^{2} + 2\cdot\frac{5}{2}\cdot x + \Bigl(\frac{5}{2}\Bigr)^{2} - \frac{13}{4}\\[5pt] | ||
| + | &= x^{2} + 5x + \frac{25}{4} - \frac{13}{4}\\[5pt] | ||
| + | &= x^{2} + 5x + \frac{12}{4}\\[5pt] | ||
| + | &= x^{2}+5x+3\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
Current revision
We apply the standard formula for completing the square,
| \displaystyle x^{2}+ax = \Bigl(x+\frac{a}{2}\Bigr)^{2} - \Bigl(\frac{a}{2}\Bigr)^{2}\,\textrm{,} |
on our expression and this gives
| \displaystyle x^{2}+5x = \Bigl(x+\frac{5}{2}\Bigr)^{2} - \Bigl(\frac{5}{2}\Bigr)^{2} = \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{25}{4}\,\textrm{.} |
The whole expression becomes
| \displaystyle \begin{align}
x^{2}+5x+3 &= \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{25}{4}+3\\[5pt] &= \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{25}{4} + \frac{12}{4}\\[5pt] &= \Bigl(x+\frac{5}{2}\Bigr)^{2} + \frac{12-25}{4}\\[5pt] &= \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{13}{4}\,\textrm{.} \end{align} |
A quick check shows that we have calculated correctly
| \displaystyle \begin{align}
\Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{13}{4} &= x^{2} + 2\cdot\frac{5}{2}\cdot x + \Bigl(\frac{5}{2}\Bigr)^{2} - \frac{13}{4}\\[5pt] &= x^{2} + 5x + \frac{25}{4} - \frac{13}{4}\\[5pt] &= x^{2} + 5x + \frac{12}{4}\\[5pt] &= x^{2}+5x+3\,\textrm{.} \end{align} |
