Solution 2.3:1b
From Förberedande kurs i matematik 1
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- | When we complete the square, it is only the first two terms, | + | When we complete the square, it is only the first two terms, <math>x^{2}+2x</math>, that are involved. The general formula for completing the square states that <math>x^{2}+ax</math> equals |
- | <math>x^{2}+2x</math> | + | |
- | , that are involved. The general | + | |
- | formula for completing the square states that | + | |
- | <math>x^{2}+ax</math> | + | |
- | equals | + | |
+ | {{Displayed math||<math>\biggl(x+\frac{a}{2}\biggr)^{2} - \biggl(\frac{a}{2}\biggr)^{2}\,\textrm{.}</math>}} | ||
- | + | Note how the coefficient ''a'' in front of the ''x'' turns up halved in two places. | |
- | + | ||
- | + | ||
- | Note how the coefficient | + | |
- | + | ||
- | in front of the | + | |
- | + | ||
- | turns up halved in two places. | + | |
If we use this formula, we obtain | If we use this formula, we obtain | ||
+ | {{Displayed math||<math>x^{2}+2x = \biggl(x+\frac{2}{2}\biggr)^{2} - \biggl(\frac{2}{2}\biggr)^{2} = (x+1)^{2}-1</math>}} | ||
- | + | and if we subtract the last "1", we obtain | |
- | + | ||
- | + | ||
- | and if we subtract the last " | + | |
- | + | ||
- | " , we obtain | + | |
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+ | {{Displayed math||<math>x^{2}+2x-1 = (x+1)^{2}-1-1 = (x+1)^{2}-2\,\textrm{.}</math>}} | ||
To be completely certain that we have used the correct formula, we can expand the quadratic on the right-hand side, | To be completely certain that we have used the correct formula, we can expand the quadratic on the right-hand side, | ||
- | + | {{Displayed math||<math>(x+1)^{2}-2 = x^{2}+2x+1-2 = x^{2}+2x-1</math>}} | |
- | <math> | + | |
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and see that the relation really holds. | and see that the relation really holds. |
Revision as of 13:39, 26 September 2008
When we complete the square, it is only the first two terms, \displaystyle x^{2}+2x, that are involved. The general formula for completing the square states that \displaystyle x^{2}+ax equals
\displaystyle \biggl(x+\frac{a}{2}\biggr)^{2} - \biggl(\frac{a}{2}\biggr)^{2}\,\textrm{.} |
Note how the coefficient a in front of the x turns up halved in two places.
If we use this formula, we obtain
\displaystyle x^{2}+2x = \biggl(x+\frac{2}{2}\biggr)^{2} - \biggl(\frac{2}{2}\biggr)^{2} = (x+1)^{2}-1 |
and if we subtract the last "1", we obtain
\displaystyle x^{2}+2x-1 = (x+1)^{2}-1-1 = (x+1)^{2}-2\,\textrm{.} |
To be completely certain that we have used the correct formula, we can expand the quadratic on the right-hand side,
\displaystyle (x+1)^{2}-2 = x^{2}+2x+1-2 = x^{2}+2x-1 |
and see that the relation really holds.