Lösung 3.2:3

Aus Online Mathematik Brückenkurs 1

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First, we move the
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First, we move the 2 to the right-hand side to get <math>\sqrt{3x-8}=x-2</math> and then square away the root sign,
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<math>\text{2}</math>
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to the right-hand side to get
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<math>\sqrt{3x-8}=x-2</math>
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and then square away the root sign,
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(*)
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{{Displayed math||<math>3x-8 = (x-2)^{2}</math>|(*)}}
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<math>3x-8=\left( x-2 \right)^{2}</math>
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or, with the right-hand side expanded
or, with the right-hand side expanded
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{{Displayed math||<math>3x-8=x^{2}-4x+4\,\textrm{.}</math>}}
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<math>3x-8=x^{2}-4x+4</math>
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If we move over all the terms to the left-hand side, we get
If we move over all the terms to the left-hand side, we get
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{{Displayed math||<math>x^{2}-7x+12=0\,\textrm{.}</math>}}
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<math>x^{2}-7x+12=0</math>
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If we complete the square of the left-hand side,
If we complete the square of the left-hand side,
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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x^2-7x+12 &= \Bigl(x-\frac{7}{2}\Bigr)^2 - \Bigl(\frac{7}{2}\Bigr)^2 + 12\\[5pt]
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& x^{2}-7x+12=\left( x-\frac{7}{2} \right)^{2}-\left( \frac{7}{2} \right)^{2}+12 \\
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&= \Bigl(x-\frac{7}{2}\Bigr)^2 - \frac{49}{4} + \frac{48}{4}\\[5pt]
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& =\left( x-\frac{7}{2} \right)^{2}-\frac{49}{4}+\frac{48}{4} \\
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&= \Bigl(x-\frac{7}{2}\Bigr)^2 - \frac{1}{4}
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& =\left( x-\frac{7}{2} \right)^{2}-\frac{1}{4} \\
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\end{align}</math>}}
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\end{align}</math>
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the equation can be written as
the equation can be written as
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{{Displayed math||<math>\Bigl(x-\frac{7}{2}\Bigr)^{2} = \frac{1}{4}</math>}}
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<math>\left( x-\frac{7}{2} \right)^{2}=\frac{1}{4}</math>
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and the solutions are
and the solutions are
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:*<math>x = \frac{7}{2} + \sqrt{\frac{1}{4}} = \frac{7}{2} + \frac{1}{2} = \frac{8}{2} = 4\,,</math>
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<math>x=\frac{7}{2}+\sqrt{\frac{1}{4}}=\frac{7}{2}+\frac{1}{2}=\frac{8}{2}=4</math>
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:*<math>x = \frac{7}{2} - \sqrt{\frac{1}{4}} = \frac{7}{2} - \frac{1}{2} = \frac{6}{2} = 3\,\textrm{.}</math>
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<math>x=\frac{7}{2}-\sqrt{\frac{1}{4}}=\frac{7}{2}-\frac{1}{2}=\frac{6}{2}=3</math>
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To be on the safe side, we verify that
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<math>x=\text{3 }</math>
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and
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<math>x=\text{4}</math>
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satisfy the squared equation (*)
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<math>x=\text{3 }</math>: LHS
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<math>=3\centerdot 3-8=9-8=1</math>
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and RHS
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<math>=\left( 3-2 \right)^{2}=1</math>
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<math>x=\text{4}</math>: LHS
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<math>=3\centerdot 4-8=12-8=4</math>
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and RHS
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<math>=\left( 4\centerdot 2 \right)^{2}=4</math>
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Because we squared the root equation, possible false roots turn up and we therefore have to verify the solutions when we go back to the original root equation:
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<math>x=\text{3 }</math>: LHS
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To be on the safe side, we verify that <math>x=3</math> and <math>x=4</math> satisfy the squared equation (*)
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<math>=\sqrt{3\centerdot 3-8}+2=\sqrt{9-8}+2=1+2=3</math>
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RHS
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<math>=3</math>
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{|
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||<ul><li>''x''&nbsp;=&nbsp;3:</li></ul>
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||<math>\ \text{LHS} = 3\cdot 3-8 = 9-8 = 1</math> and
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|-
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||
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||<math>\ \text{RHS} = (3-2)^2 = 1</math>
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|-
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||<ul><li>''x''&nbsp;=&nbsp;4:</li></ul>
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||<math>\ \text{LHS} = 3\cdot 4-8 = 12-8 = 4</math> and
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|-
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||
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||<math>\ \text{RHS} = (4-2)^2 = 4</math>
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|}
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<math>x=\text{4}</math>: LHS
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Because we squared the root equation, possible spurious roots turn up and we therefore have to verify the solutions when we go back to the original root equation:
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<math>=\sqrt{3\centerdot 4-8}-2=\sqrt{12-8}-2=2+2=4</math>
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RHS
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<math>=4</math>
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{|
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||<ul><li>''x''&nbsp;=&nbsp;3:</li></ul>
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||<math>\ \text{LHS} = \sqrt{3\cdot 3-8} + 2 = \sqrt{9-8} + 2 = 1+2 = 3</math> and
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|-
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||<math>\ \text{RHS} = 3</math>
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|-
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||<ul><li>''x''&nbsp;=&nbsp;4:</li></ul>
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||<math>\ \text{LHS} = \sqrt{3\cdot 4-8}-2 = \sqrt{12-8}+2 = 2+2 = 4</math> and
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|-
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||<math>\ \text{RHS} = 4</math>
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|}
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The solutions to the root equation are
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The solutions to the root equation are <math>x=3</math> and <math>x=4</math>.
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<math>x=\text{3 }</math>
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and
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<math>x=\text{4}</math>.
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Version vom 12:06, 1. Okt. 2008

First, we move the 2 to the right-hand side to get \displaystyle \sqrt{3x-8}=x-2 and then square away the root sign,

Vorlage:Displayed math

or, with the right-hand side expanded

Vorlage:Displayed math

If we move over all the terms to the left-hand side, we get

Vorlage:Displayed math

If we complete the square of the left-hand side,

Vorlage:Displayed math

the equation can be written as

Vorlage:Displayed math

and the solutions are

  • \displaystyle x = \frac{7}{2} + \sqrt{\frac{1}{4}} = \frac{7}{2} + \frac{1}{2} = \frac{8}{2} = 4\,,
  • \displaystyle x = \frac{7}{2} - \sqrt{\frac{1}{4}} = \frac{7}{2} - \frac{1}{2} = \frac{6}{2} = 3\,\textrm{.}

To be on the safe side, we verify that \displaystyle x=3 and \displaystyle x=4 satisfy the squared equation (*)

  • x = 3:
\displaystyle \ \text{LHS} = 3\cdot 3-8 = 9-8 = 1 and
\displaystyle \ \text{RHS} = (3-2)^2 = 1
  • x = 4:
\displaystyle \ \text{LHS} = 3\cdot 4-8 = 12-8 = 4 and
\displaystyle \ \text{RHS} = (4-2)^2 = 4

Because we squared the root equation, possible spurious roots turn up and we therefore have to verify the solutions when we go back to the original root equation:

  • x = 3:
\displaystyle \ \text{LHS} = \sqrt{3\cdot 3-8} + 2 = \sqrt{9-8} + 2 = 1+2 = 3 and
\displaystyle \ \text{RHS} = 3
  • x = 4:
\displaystyle \ \text{LHS} = \sqrt{3\cdot 4-8}-2 = \sqrt{12-8}+2 = 2+2 = 4 and
\displaystyle \ \text{RHS} = 4

The solutions to the root equation are \displaystyle x=3 and \displaystyle x=4.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.2:3“