Solution 2.3:1c

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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
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<math>2x-x^{2}</math>, which we also can write as
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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
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<math>-\left( x^{2}-2x \right)</math>
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. If we neglect the minus sign, we can complete square of the expression
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<math>2x-x^{2}</math>
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by using the formula
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<math>x^{2}-ax=\left( x-\frac{a}{2} \right)^{2}-\left( \frac{a}{2} \right)^{2}</math>
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{{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
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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\,\textrm{.}</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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This means that
This means that
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{{Displayed math||<math>\begin{align}
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5+2x-x^{2} &= 5-(x^{2}-2x) = 5-\bigl((x-1)^{2}-1\bigr)\\[5pt]
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&= 5-(x-1)^{2}+1 = 6-(x-1)^{2}\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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A quick check shows that we have completed the square correctly
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& 5+2x-x^{2}=5-\left( x^{2}-2x \right)=5-\left( \left( x-1 \right)^{2}-1 \right) \\
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& \\
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& =5-\left( x-1 \right)^{2}+1=6-\left( x-1 \right)^{2} \\
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& \\
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\end{align}</math>
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A quick check shows that we have completed the square correctly.:
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<math>\begin{align}
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{{Displayed math||<math>\begin{align}
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& 6-\left( x-1 \right)^{2}=6-\left( x^{2}-2x+1 \right)=6-x^{2}+2x-1 \\
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6-(x-1)^{2}
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& \\
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&= 6-(x^{2}-2x+1)\\[5pt]
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& =5+2x-x^{2} \\
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&= 6-x^{2}+2x-1\\[5pt]
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& \\
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& =5+2x-x^{2}\textrm{.}
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\end{align}</math>
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\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}

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