Solution 2.3:2f

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Current revision (08:24, 29 September 2008) (edit) (undo)
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We divide both sides by
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We divide both sides by 3 and complete the square on the left-hand side,
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<math>3</math>
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and complete the square on the left-hand side:
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<math>\begin{align}
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& x^{2}-\frac{10}{3}x+\frac{8}{3}=\left( x-\frac{5}{3} \right)^{2}-\left( -\frac{5}{3} \right)^{2}+\frac{8}{3}=\left( x-\frac{5}{3} \right)^{2}-\frac{25}{9}+\frac{24}{9} \\
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& =\left( x-\frac{5}{3} \right)^{2}-\frac{1}{9} \\
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\end{align}</math>
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{{Displayed math||<math>\begin{align}
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x^{2}-\frac{10}{3}x+\frac{8}{3}
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&= \Bigl(x-\frac{5}{3}\Bigr)^{2} - \Bigl(\frac{5}{3}\Bigr)^{2} + \frac{8}{3}\\[5pt]
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&= \Bigl(x-\frac{5}{3}\Bigr)^{2} - \frac{25}{9} + \frac{24}{9}\\[5pt]
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&= \Bigl(x-\frac{5}{3}\Bigr)^{2} - \frac{1}{9}\,\textrm{.}
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\end{align}</math>}}
The equation then becomes
The equation then becomes
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{{Displayed math||<math>\left( x-\frac{5}{3} \right)^{2}=\frac{1}{9}\,\textrm{,}</math>}}
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<math>\left( x-\frac{5}{3} \right)^{2}=\frac{1}{9}</math>
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and taking the square root gives the solutions as
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and taking the root gives the solutions as
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:*<math>x-\tfrac{5}{3} = \sqrt{\tfrac{1}{9}} = \tfrac{1}{3}\,,\quad</math> i.e. <math>x = \tfrac{5}{3} + \tfrac{1}{3} = \tfrac{6}{3} = 2\,\textrm{,}</math>
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<math>x-\frac{5}{3}=\sqrt{\frac{1}{9}}=\frac{1}{3}</math>
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i.e.
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<math>x=\frac{5}{3}+\frac{1}{3}=\frac{6}{3}=2.</math>
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<math>x-\frac{5}{3}=-\sqrt{\frac{1}{9}}=-\frac{1}{3}</math>
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i.e.
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<math>x=\frac{5}{3}-\frac{1}{3}=\frac{4}{3}.</math>
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:*<math>x-\tfrac{5}{3} = -\sqrt{\tfrac{1}{9}} = -\tfrac{1}{3}\,,\quad</math> i.e. <math>x = \tfrac{5}{3} - \tfrac{1}{3} = \tfrac{4}{3}\,\textrm{.}</math>
Check:
Check:
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:*''x''&nbsp;=&nbsp;4/3: <math>\ \text{LHS} = 3\cdot\bigl(\tfrac{4}{3}\bigr)^{2} - 10\cdot\tfrac{4}{3} + 8 = 3\cdot\tfrac{16}{9} - \tfrac{40}{3} + \tfrac{8\cdot 3}{3} = 0 = \text{RHS,}</math>
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<math>x=\text{4}/\text{3}</math>: LHS
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:*''x''&nbsp;=&nbsp;2: <math>\ \text{LHS} = 3\cdot 2^{2} - 10\cdot 2 + 8 = 12 - 20 + 8 = 0 = \text{RHS.}</math>
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<math>=3\centerdot \left( \frac{4}{3} \right)^{2}-10\centerdot \frac{4}{3}+8=3\centerdot \frac{16}{9}-\frac{40}{3}+\frac{8\centerdot 3}{3}=\frac{16-40+24}{3}=0=</math>
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RHS
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<math>x=\text{2}</math>: LHS
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<math>=3\centerdot 2^{2}-10\centerdot 2+8=12-20+8=0=</math>
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RHS
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Current revision

We divide both sides by 3 and complete the square on the left-hand side,

\displaystyle \begin{align}

x^{2}-\frac{10}{3}x+\frac{8}{3} &= \Bigl(x-\frac{5}{3}\Bigr)^{2} - \Bigl(\frac{5}{3}\Bigr)^{2} + \frac{8}{3}\\[5pt] &= \Bigl(x-\frac{5}{3}\Bigr)^{2} - \frac{25}{9} + \frac{24}{9}\\[5pt] &= \Bigl(x-\frac{5}{3}\Bigr)^{2} - \frac{1}{9}\,\textrm{.} \end{align}

The equation then becomes

\displaystyle \left( x-\frac{5}{3} \right)^{2}=\frac{1}{9}\,\textrm{,}

and taking the square root gives the solutions as

  • \displaystyle x-\tfrac{5}{3} = \sqrt{\tfrac{1}{9}} = \tfrac{1}{3}\,,\quad i.e. \displaystyle x = \tfrac{5}{3} + \tfrac{1}{3} = \tfrac{6}{3} = 2\,\textrm{,}
  • \displaystyle x-\tfrac{5}{3} = -\sqrt{\tfrac{1}{9}} = -\tfrac{1}{3}\,,\quad i.e. \displaystyle x = \tfrac{5}{3} - \tfrac{1}{3} = \tfrac{4}{3}\,\textrm{.}

Check:

  • x = 4/3: \displaystyle \ \text{LHS} = 3\cdot\bigl(\tfrac{4}{3}\bigr)^{2} - 10\cdot\tfrac{4}{3} + 8 = 3\cdot\tfrac{16}{9} - \tfrac{40}{3} + \tfrac{8\cdot 3}{3} = 0 = \text{RHS,}
  • x = 2: \displaystyle \ \text{LHS} = 3\cdot 2^{2} - 10\cdot 2 + 8 = 12 - 20 + 8 = 0 = \text{RHS.}
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