Solution 2.3:1d

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We apply the standard formula for completing the square,
We apply the standard formula for completing the square,
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{{Displayed math||<math>x^{2}+ax = \Bigl(x+\frac{a}{2}\Bigr)^{2} - \Bigl(\frac{a}{2}\Bigr)^{2}\,\textrm{,}</math>}}
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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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on our expression and this gives
on our expression and this gives
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{{Displayed math||<math>x^{2}+5x = \Bigl(x+\frac{5}{2}\Bigr)^{2} - \Bigl(\frac{5}{2}\Bigr)^{2} = \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{25}{4}\,\textrm{.}</math>}}
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<math>x^{2}+5x=\left( x+\frac{5}{2} \right)^{2}-\left( \frac{5}{2} \right)^{2}=\left( x+\frac{5}{2} \right)^{2}-\frac{25}{4}</math>
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The whole expression becomes
The whole expression becomes
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{{Displayed math||<math>\begin{align}
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x^{2}+5x+3
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&= \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{25}{4}+3\\[5pt]
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&= \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{25}{4} + \frac{12}{4}\\[5pt]
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&= \Bigl(x+\frac{5}{2}\Bigr)^{2} + \frac{12-25}{4}\\[5pt]
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&= \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{13}{4}\,\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 calculated correctly
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& x^{2}+5x+3=\left( x+\frac{5}{2} \right)^{2}-\frac{25}{4}+3=\left( x+\frac{5}{2} \right)^{2}-\frac{25}{4}+\frac{12}{4} \\
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& =\left( x+\frac{5}{2} \right)^{2}+\frac{12-25}{4}=\left( x+\frac{5}{2} \right)^{2}-\frac{13}{4} \\
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\end{align}</math>
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A quick check shows that we have calculated correctly.
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<math>\begin{align}
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{{Displayed math||<math>\begin{align}
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& \left( x+\frac{5}{2} \right)^{2}-\frac{13}{4}=x^{2}+2\centerdot \frac{5}{2}\centerdot x+\left( \frac{5}{2} \right)^{2}-\frac{13}{4}=x^{2}+5x+\frac{25}{4}-\frac{13}{4} \\
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\Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{13}{4}
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& =x^{2}+5x+\frac{12}{4}=x^{2}+5x+3 \\
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&= x^{2} + 2\cdot\frac{5}{2}\cdot x + \Bigl(\frac{5}{2}\Bigr)^{2} - \frac{13}{4}\\[5pt]
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\end{align}</math>
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&= x^{2} + 5x + \frac{25}{4} - \frac{13}{4}\\[5pt]
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&= x^{2} + 5x + \frac{12}{4}\\[5pt]
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&= x^{2}+5x+3\,\textrm{.}
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\end{align}</math>}}

Current revision

We apply the standard formula for completing the square,

\displaystyle x^{2}+ax = \Bigl(x+\frac{a}{2}\Bigr)^{2} - \Bigl(\frac{a}{2}\Bigr)^{2}\,\textrm{,}

on our expression and this gives

\displaystyle x^{2}+5x = \Bigl(x+\frac{5}{2}\Bigr)^{2} - \Bigl(\frac{5}{2}\Bigr)^{2} = \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{25}{4}\,\textrm{.}

The whole expression becomes

\displaystyle \begin{align}

x^{2}+5x+3 &= \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{25}{4}+3\\[5pt] &= \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{25}{4} + \frac{12}{4}\\[5pt] &= \Bigl(x+\frac{5}{2}\Bigr)^{2} + \frac{12-25}{4}\\[5pt] &= \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{13}{4}\,\textrm{.} \end{align}

A quick check shows that we have calculated correctly

\displaystyle \begin{align}

\Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{13}{4} &= x^{2} + 2\cdot\frac{5}{2}\cdot x + \Bigl(\frac{5}{2}\Bigr)^{2} - \frac{13}{4}\\[5pt] &= x^{2} + 5x + \frac{25}{4} - \frac{13}{4}\\[5pt] &= x^{2} + 5x + \frac{12}{4}\\[5pt] &= x^{2}+5x+3\,\textrm{.} \end{align}

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