Solution 2.3:1c
From Förberedande kurs i matematik 1
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As always when completing the square, we focus on the quadratic and linear terms | As always when completing the square, we focus on the quadratic and linear terms | ||
- | <math>2x-x^{2}</math> | + | <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 |
- | , which we also can write as | + | |
- | <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 = \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 = \Bigl(x-\frac{2}{2}\Bigr)^{2} - \Bigl(\frac{2}{2}\Bigr)^{2} = (x-1)^{2}-1\,\textrm{.}</math>}} | |
- | <math>x^{2}-2x=\ | + | |
- | + | ||
This means that | This means that | ||
+ | {{Displayed math||<math>\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}</math>}} | ||
- | + | A quick check shows that we have completed the square correctly | |
- | + | ||
- | + | ||
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- | + | ||
- | + | ||
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- | A quick check shows that we have completed the square correctly | + | |
- | + | ||
- | <math>\begin{align} | + | {{Displayed math||<math>\begin{align} |
- | + | 6-(x-1)^{2} | |
- | + | &= 6-(x^{2}-2x+1)\\[5pt] | |
- | & =5+2x-x^{2} \ | + | &= 6-x^{2}+2x-1\\[5pt] |
- | + | & =5+2x-x^{2}\textrm{.} | |
- | \end{align}</math> | + | \end{align}</math>}} |
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} |