Solution 3.2:5

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After squaring both sides, we obtain the equation
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<center> [[Bild:3_2_5-1(2).gif]] </center>
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{{Displayed math||<math>3x-2 = (2-x)^2</math>|(*)}}
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<center> [[Bild:3_2_5-2(2).gif]] </center>
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and if we expand the right-hand side and then collect the terms, we get
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{{Displayed math||<math>x^{2}-7x+6=0\,\textrm{.}</math>}}
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Completing the square of the left-hand side, we obtain
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{{Displayed math||<math>\begin{align}
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x^{2}-7x+6
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&= \Bigl(x-\frac{7}{2}\Bigr)^2 - \Bigl(\frac{7}{2}\Bigr)^2+6\\[5pt]
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&= \Bigl(x-\frac{7}{2}\Bigr)^2 - \frac{49}{4} + \frac{24}{4}\\[5pt]
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&= \Bigl(x-\frac{7}{2}\Bigr)^2 - \frac{25}{4}
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\end{align}</math>}}
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which means that the equation can be written as
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{{Displayed math||<math>\Bigl(x-\frac{7}{2}\Bigr)^2 = \frac{25}{4}</math>}}
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and the solutions are therefore
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:*<math>x = \frac{7}{2} + \sqrt{\frac{25}{4}} = \frac{7}{2} + \frac{5}{2} = \frac{12}{2} = 6\,,</math>
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:*<math>x = \frac{7}{2} - \sqrt{\frac{25}{4}} = \frac{7}{2} - \frac{5}{2} = \frac{2}{2} = 1\,\textrm{.}</math>
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Substituting <math>x=1</math> and <math>x=6</math> into the quadratic equation (*) shows that we have solved the equation correctly.
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:*''x''&nbsp;=&nbsp;1: <math>\ \text{LHS} = 3\cdot 1-2 = 1\ </math> and <math>\ \text{RHS} = (2-1)^2 = 1</math>
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:*''x''&nbsp;=&nbsp;6: <math>\ \text{LHS} = 3\cdot 6-2 = 16\ </math> and <math>\ \text{RHS} = (2-6)^2 = 16</math>
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Finally, we need to sort away possible spurious roots to the root equation by verifying the solutions.
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:*''x''&nbsp;=&nbsp;1: <math>\ \text{LHS} = \sqrt{3\cdot 1-2} = 1\ </math> and <math>\ \text{RHS} = 2-1 = 1</math>
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:*''x''&nbsp;=&nbsp;6: <math>\ \text{LHS} = \sqrt{3\cdot 6-2} = 4\ </math> and <math>\ \text{RHS} = 2-6 = -4</math>
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This shows that the root equation has the solution <math>x=1\,</math>.

Current revision

After squaring both sides, we obtain the equation

\displaystyle 3x-2 = (2-x)^2 (*)

and if we expand the right-hand side and then collect the terms, we get

\displaystyle x^{2}-7x+6=0\,\textrm{.}

Completing the square of the left-hand side, we obtain

\displaystyle \begin{align}

x^{2}-7x+6 &= \Bigl(x-\frac{7}{2}\Bigr)^2 - \Bigl(\frac{7}{2}\Bigr)^2+6\\[5pt] &= \Bigl(x-\frac{7}{2}\Bigr)^2 - \frac{49}{4} + \frac{24}{4}\\[5pt] &= \Bigl(x-\frac{7}{2}\Bigr)^2 - \frac{25}{4} \end{align}

which means that the equation can be written as

\displaystyle \Bigl(x-\frac{7}{2}\Bigr)^2 = \frac{25}{4}

and the solutions are therefore

  • \displaystyle x = \frac{7}{2} + \sqrt{\frac{25}{4}} = \frac{7}{2} + \frac{5}{2} = \frac{12}{2} = 6\,,
  • \displaystyle x = \frac{7}{2} - \sqrt{\frac{25}{4}} = \frac{7}{2} - \frac{5}{2} = \frac{2}{2} = 1\,\textrm{.}

Substituting \displaystyle x=1 and \displaystyle x=6 into the quadratic equation (*) shows that we have solved the equation correctly.

  • x = 1: \displaystyle \ \text{LHS} = 3\cdot 1-2 = 1\ and \displaystyle \ \text{RHS} = (2-1)^2 = 1
  • x = 6: \displaystyle \ \text{LHS} = 3\cdot 6-2 = 16\ and \displaystyle \ \text{RHS} = (2-6)^2 = 16

Finally, we need to sort away possible spurious roots to the root equation by verifying the solutions.

  • x = 1: \displaystyle \ \text{LHS} = \sqrt{3\cdot 1-2} = 1\ and \displaystyle \ \text{RHS} = 2-1 = 1
  • x = 6: \displaystyle \ \text{LHS} = \sqrt{3\cdot 6-2} = 4\ and \displaystyle \ \text{RHS} = 2-6 = -4

This shows that the root equation has the solution \displaystyle x=1\,.

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