Solution 2.3:1b

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When we complete the square, it is only the first two terms,
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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
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<math>x^{2}+2x</math>
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, that are involved. The general
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formula for completing the square states that
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<math>x^{2}+ax</math>
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equals
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{{Displayed math||<math>\Bigl(x+\frac{a}{2}\Bigr)^{2} - \Bigl(\frac{a}{2}\Bigr)^{2}\,\textrm{.}</math>}}
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<math>\left( x+\frac{a}{2} \right)^{2}-\left( \frac{a}{2} \right)^{2}</math>
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Note how the coefficient ''a'' in front of the ''x'' turns up halved in two places.
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Note how the coefficient
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<math>a</math>
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in front of the
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<math>x</math>
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turns up halved in two places.
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If we use this formula, we obtain
If we use this formula, we obtain
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{{Displayed math||<math>x^{2}+2x = \Bigl(x+\frac{2}{2}\Bigr)^{2} - \Bigl(\frac{2}{2}\Bigr)^{2} = (x+1)^{2}-1</math>}}
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<math>x^{2}+2x=\left( x+\frac{2}{2} \right)^{2}-\left( \frac{2}{2} \right)^{2}=\left( x+1 \right)^{2}-1</math>
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and if we subtract the last "1", we obtain
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and if we subtract the last "
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<math>1</math>
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" , we obtain
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<math>x^{2}+2x-1=\left( x+1 \right)^{2}-1-1=\left( x+1 \right)^{2}-2</math>
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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,
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{{Displayed math||<math>(x+1)^{2}-2 = x^{2}+2x+1-2 = x^{2}+2x-1</math>}}
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<math>\left( x+1 \right)^{2}-2=x^{2}+2x+1-2=x^{2}+2x-1</math>
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and see that the relation really holds.
and see that the relation really holds.

Current revision

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 \Bigl(x+\frac{a}{2}\Bigr)^{2} - \Bigl(\frac{a}{2}\Bigr)^{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 = \Bigl(x+\frac{2}{2}\Bigr)^{2} - \Bigl(\frac{2}{2}\Bigr)^{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.

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