Solution 2.3:2d

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The equation can be written in normalized form (i.e. the coefficient in front of
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The equation can be written in normalized form (i.e. the coefficient in front of ''x''² is 1) by dividing both sides by 4,
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<math>x^{\text{2}}</math>
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is
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<math>1</math>
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) by dividing both sides by
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<math>4</math>,
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{{Displayed math||<math>x^{2}-7x+\frac{13}{4}=0\,\textrm{.}</math>}}
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<math>x^{2}-7x+\frac{13}{4}=0</math>
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Complete the square on the left-hand side,
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{{Displayed math||<math>\begin{align}
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Completing the square on the left-hand side,
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x^{2}-7x+\frac{13}{4}
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&= \Bigl(x-\frac{7}{2}\Bigr)^{2} - \Bigl(\frac{7}{2}\Bigr)^{2} + \frac{13}{4}\\[5pt]
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&= \Bigl(x-\frac{7}{2}\Bigr)^{2} - \frac{49}{4} + \frac{13}{4}\\[5pt]
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<math>\begin{align}
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&= \Bigl(x-\frac{7}{2}\Bigr)^{2} - \frac{36}{4}\\[5pt]
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& x^{2}-7x+\frac{13}{4}=\left( x-\frac{7}{2} \right)^{2}-\left( \frac{7}{2} \right)^{2}+\frac{13}{4}=\left( x-\frac{7}{2} \right)^{2}-\frac{49}{4}+\frac{13}{4} \\
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&= \Bigl(x-\frac{7}{2}\Bigr)^{2} - 9\,\textrm{.}
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& =\left( x-\frac{7}{2} \right)^{2}-\frac{36}{4}=\left( x-\frac{7}{2} \right)^{2}-9 \\
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\end{align}</math>}}
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\end{align}</math>
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The equation can therefore be written as
The equation can therefore be written as
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{{Displayed math||<math>\Bigl(x-\frac{7}{2}\Bigr)^{2} - 9 = 0\,,</math>}}
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<math>\left( x-\frac{7}{2} \right)^{2}-9=0</math>
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and taking the square root gives the solutions as
and taking the square root gives the solutions as
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:*<math>x-\frac{7}{2}=\sqrt{9}=3\,,\quad</math> i.e. <math>x=\frac{7}{2}+3=\frac{13}{2},</math>
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<math>x-\frac{7}{2}=\sqrt{9}=3</math>
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:*<math>x-\frac{7}{2}=-\sqrt{9}=-3\,,\quad</math> i.e. <math>x=\frac{7}{2}-3=\frac{1}{2}.</math>
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i.e.
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<math>x=\frac{7}{2}+3=\frac{13}{2},</math>
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<math>x-\frac{7}{2}=-\sqrt{9}=-3</math>
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i.e.
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<math>x=\frac{7}{2}-3=\frac{1}{2}.</math>
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As an extra check, we substitute x=1/2 and x=13/2 into the equation:
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As an extra check, we substitute ''x''&nbsp;=&nbsp;1/2 and ''x''&nbsp;=&nbsp;13/2 into the equation:
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<math>x=\text{1}/\text{2}</math>: LHS
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:*''x''&nbsp;=&nbsp;1/2: <math>\ \text{LHS} = 4\cdot\bigl(\tfrac{1}{2}\bigr)^{2} - 28\cdot\tfrac{1}{2}+13 = 4\cdot\tfrac{1}{4}-14+13 = \text{RHS,}</math>
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<math>=4\centerdot \left( \frac{1}{2} \right)^{2}-28\centerdot \frac{1}{2}+13=4\centerdot \frac{1}{4}-14+13=1-14+13=</math>
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RHS
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<math>x=\text{13}/\text{2}</math>: LHS
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:*''x''&nbsp;=&nbsp;13/2: <math>\ \text{LHS} = 4\cdot\bigl(\tfrac{13}{2}\bigr)^{2} - 28\cdot\tfrac{13}{2}+13 = 4\cdot\tfrac{169}{4} - 14\cdot 13 + 13 = \text{RHS.}</math>
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<math>=4\centerdot \left( \frac{13}{2} \right)^{2}-28\centerdot \frac{13}{2}+13=4\centerdot \frac{169}{4}-14\centerdot 13+13=169-182+13=</math>
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RHS
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Current revision

The equation can be written in normalized form (i.e. the coefficient in front of x² is 1) by dividing both sides by 4,

\displaystyle x^{2}-7x+\frac{13}{4}=0\,\textrm{.}

Complete the square on the left-hand side,

\displaystyle \begin{align}

x^{2}-7x+\frac{13}{4} &= \Bigl(x-\frac{7}{2}\Bigr)^{2} - \Bigl(\frac{7}{2}\Bigr)^{2} + \frac{13}{4}\\[5pt] &= \Bigl(x-\frac{7}{2}\Bigr)^{2} - \frac{49}{4} + \frac{13}{4}\\[5pt] &= \Bigl(x-\frac{7}{2}\Bigr)^{2} - \frac{36}{4}\\[5pt] &= \Bigl(x-\frac{7}{2}\Bigr)^{2} - 9\,\textrm{.} \end{align}

The equation can therefore be written as

\displaystyle \Bigl(x-\frac{7}{2}\Bigr)^{2} - 9 = 0\,,

and taking the square root gives the solutions as

  • \displaystyle x-\frac{7}{2}=\sqrt{9}=3\,,\quad i.e. \displaystyle x=\frac{7}{2}+3=\frac{13}{2},
  • \displaystyle x-\frac{7}{2}=-\sqrt{9}=-3\,,\quad i.e. \displaystyle x=\frac{7}{2}-3=\frac{1}{2}.

As an extra check, we substitute x = 1/2 and x = 13/2 into the equation:

  • x = 1/2: \displaystyle \ \text{LHS} = 4\cdot\bigl(\tfrac{1}{2}\bigr)^{2} - 28\cdot\tfrac{1}{2}+13 = 4\cdot\tfrac{1}{4}-14+13 = \text{RHS,}
  • x = 13/2: \displaystyle \ \text{LHS} = 4\cdot\bigl(\tfrac{13}{2}\bigr)^{2} - 28\cdot\tfrac{13}{2}+13 = 4\cdot\tfrac{169}{4} - 14\cdot 13 + 13 = \text{RHS.}
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