Solution 2.3:1c
From Förberedande kurs i matematik 1
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<math>2x-x^{2}</math>, which we also can write as <math>-(x^{2}-2x)</math>. If we neglect the minus sign, we can complete square of the expression <math>2x-x^{2}</math> by using the formula | <math>2x-x^{2}</math>, which we also can write as <math>-(x^{2}-2x)</math>. If we neglect the minus sign, we can complete square of the expression <math>2x-x^{2}</math> by using the formula | ||
| - | {{Displayed math||<math>x^{2}-ax = \ | + | {{Displayed math||<math>x^{2}-ax = \Bigl(x-\frac{a}{2}\Bigr)^{2} - \Bigl(\frac{a}{2}\Bigr)^{2}</math>}} |
and we obtain | and we obtain | ||
| - | {{Displayed math||<math>x^{2}-2x = \ | + | {{Displayed math||<math>x^{2}-2x = \Bigl(x-\frac{2}{2}\Bigr)^{2} - \Bigl(\frac{2}{2}\Bigr)^{2} = (x-1)^{2}-1\,\textrm{.}</math>}} |
This means that | This means that | ||
Current revision
As always when completing the square, we focus on the quadratic and linear terms \displaystyle 2x-x^{2}, which we also can write as \displaystyle -(x^{2}-2x). If we neglect the minus sign, we can complete square of the expression \displaystyle 2x-x^{2} by using the formula
| \displaystyle x^{2}-ax = \Bigl(x-\frac{a}{2}\Bigr)^{2} - \Bigl(\frac{a}{2}\Bigr)^{2} |
and we obtain
| \displaystyle x^{2}-2x = \Bigl(x-\frac{2}{2}\Bigr)^{2} - \Bigl(\frac{2}{2}\Bigr)^{2} = (x-1)^{2}-1\,\textrm{.} |
This means that
| \displaystyle \begin{align}
5+2x-x^{2} &= 5-(x^{2}-2x) = 5-\bigl((x-1)^{2}-1\bigr)\\[5pt] &= 5-(x-1)^{2}+1 = 6-(x-1)^{2}\textrm{.} \end{align} |
A quick check shows that we have completed the square correctly
| \displaystyle \begin{align}
6-(x-1)^{2} &= 6-(x^{2}-2x+1)\\[5pt] &= 6-x^{2}+2x-1\\[5pt] & =5+2x-x^{2}\textrm{.} \end{align} |
