Solution 2.3:2e

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Write the equation in normalized form by dividing both sides by 5,
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<center> [[Image:2_3_2e.gif]] </center>
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{{Displayed math||<math>x^{2}+\frac{2}{5}x-\frac{3}{5}=0\,\textrm{.}</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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x^{2}+\frac{2}{5}x-\frac{3}{5}
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&= \Bigl(x+\frac{2/5}{2}\Bigr)^{2} - \Bigl(\frac{2/5}{2}\Bigr)^{2} - \frac{3}{5}\\[5pt]
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&= \Bigl(x+\frac{1}{5}\Bigr)^{2} - \Bigl(\frac{1}{5}\Bigr)^{2} - \frac{3}{5}\\[5pt]
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&= \Bigl(x+\frac{1}{5}\Bigr)^{2} - \frac{1}{25} - \frac{3\cdot 5}{25}\\[5pt]
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&= \Bigl(x+\frac{1}{5}\Bigr)^{2} - \frac{16}{25}\,\textrm{.}
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\end{align}</math>}}
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The equation is now rewritten as
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{{Displayed math||<math>\left( x+\frac{1}{5} \right)^{2}=\frac{16}{25}\,\textrm{,}</math>}}
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and taking the root gives the solutions
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:*<math>x+\tfrac{1}{5} = \sqrt{\tfrac{16}{25}} = \tfrac{4}{5}</math> because <math>\bigl(\tfrac{4}{5}\bigr)^{2} = \tfrac{16}{25}\,,</math> which gives <math>x=-\tfrac{1}{5}+\tfrac{4}{5}=\tfrac{3}{5},</math>
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:*<math>x+\tfrac{1}{5} = -\sqrt{\tfrac{16}{25}} = -\tfrac{4}{5}\,,</math> which gives <math>x = -\tfrac{1}{5}-\tfrac{4}{5}=-1\,\textrm{.}</math>
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Finally, we check the answer by substituting <math>x=-1</math> and <math>x=3/5</math> into the equation:
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:*''x''&nbsp;=&nbsp;1: <math>\ \text{LHS} = 5\cdot (-1)^{2} + 2\cdot (-1) - 3 = 5 - 2 - 3 = 0 = \text{RHS,}</math>
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:*''x''&nbsp;=&nbsp;3/5: <math>\ \text{LHS} = 5\cdot\bigl(\tfrac{3}{5}\bigr)^{2} + 2\cdot\bigl(\tfrac{3}{5}\bigr) - 3 = 5\cdot\tfrac{9}{25} + \tfrac{6}{5} - \tfrac{3\cdot 5}{5} = 0 = \text{RHS.}</math>

Current revision

Write the equation in normalized form by dividing both sides by 5,

\displaystyle x^{2}+\frac{2}{5}x-\frac{3}{5}=0\,\textrm{.}

Complete the square on the left-hand side,

\displaystyle \begin{align}

x^{2}+\frac{2}{5}x-\frac{3}{5} &= \Bigl(x+\frac{2/5}{2}\Bigr)^{2} - \Bigl(\frac{2/5}{2}\Bigr)^{2} - \frac{3}{5}\\[5pt] &= \Bigl(x+\frac{1}{5}\Bigr)^{2} - \Bigl(\frac{1}{5}\Bigr)^{2} - \frac{3}{5}\\[5pt] &= \Bigl(x+\frac{1}{5}\Bigr)^{2} - \frac{1}{25} - \frac{3\cdot 5}{25}\\[5pt] &= \Bigl(x+\frac{1}{5}\Bigr)^{2} - \frac{16}{25}\,\textrm{.} \end{align}

The equation is now rewritten as

\displaystyle \left( x+\frac{1}{5} \right)^{2}=\frac{16}{25}\,\textrm{,}

and taking the root gives the solutions

  • \displaystyle x+\tfrac{1}{5} = \sqrt{\tfrac{16}{25}} = \tfrac{4}{5} because \displaystyle \bigl(\tfrac{4}{5}\bigr)^{2} = \tfrac{16}{25}\,, which gives \displaystyle x=-\tfrac{1}{5}+\tfrac{4}{5}=\tfrac{3}{5},
  • \displaystyle x+\tfrac{1}{5} = -\sqrt{\tfrac{16}{25}} = -\tfrac{4}{5}\,, which gives \displaystyle x = -\tfrac{1}{5}-\tfrac{4}{5}=-1\,\textrm{.}

Finally, we check the answer by substituting \displaystyle x=-1 and \displaystyle x=3/5 into the equation:

  • x = 1: \displaystyle \ \text{LHS} = 5\cdot (-1)^{2} + 2\cdot (-1) - 3 = 5 - 2 - 3 = 0 = \text{RHS,}
  • x = 3/5: \displaystyle \ \text{LHS} = 5\cdot\bigl(\tfrac{3}{5}\bigr)^{2} + 2\cdot\bigl(\tfrac{3}{5}\bigr) - 3 = 5\cdot\tfrac{9}{25} + \tfrac{6}{5} - \tfrac{3\cdot 5}{5} = 0 = \text{RHS.}
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