Lösung 3.2:6

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Zeile 3: Zeile 3:
Squaring once gives
Squaring once gives
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{{Displayed math||<math>\bigl(\sqrt{x+1}+\sqrt{x+5}\,\bigr)^2 = 4^2</math>}}
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<math>\left( \sqrt{x+1}+\sqrt{x+5} \right)^{2}=4^{2}</math>
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and expanding the left-hand side gives
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and expanding the left-hand side with the squaring rule gives
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{{Displayed math||<math>\bigl(\sqrt{x+1}\bigr)^2 + 2\sqrt{x+1}\sqrt{x+5} + \bigl(\sqrt{x+5}\bigr)^2 = 16</math>}}
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which, after simplification, results in the equation
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<math>\left( \sqrt{x+1} \right)^{2}+2\sqrt{x+1}\sqrt{x+5}+\left( \sqrt{x+5} \right)^{2}=16</math>
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{{Displayed math||<math>x+1+2\sqrt{x+1}\sqrt{x+5}+x+5=16\,\textrm{.}</math>}}
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which, after simplification, results in the equation:
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<math>x+1+2\sqrt{x+1}\sqrt{x+5}+x+5=16</math>
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Moving all the terms, other than the root term, over to the right-hand side,
Moving all the terms, other than the root term, over to the right-hand side,
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{{Displayed math||<math>2\sqrt{x+1}\sqrt{x+5}=-2x+10</math>}}
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<math>2\sqrt{x+1}\sqrt{x+5}=-2x+10</math>
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and squaring once again,
and squaring once again,
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{{Displayed math||<math>\bigl(2\sqrt{x+1}\sqrt{x+5}\bigr)^2 = (-2x+10)^2</math>}}
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<math>\left( 2\sqrt{x+1}\sqrt{x+5} \right)^{2}=\left( -2x+10 \right)^{2}</math>
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at last gives an equation that is completely free of root signs,
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at last gives an equation that is completely free of root signs:
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<math>4\left( x+1 \right)\left( x+5 \right)=\left( -2x+10 \right)^{2}</math>
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{{Displayed math||<math>4(x+1)(x+5) = (-2x+10)^{2}\,\textrm{.}</math>}}
Expand both sides
Expand both sides
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{{Displayed math||<math>4(x^{2}+6x+5) = 4x^2-40x+100</math>}}
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<math>4\left( x^{2}+6x+5 \right)=4x^{2}-40x+100</math>
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and then cancel the common ''x''²-term,
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{{Displayed math||<math>24x+20=-40x+100\,\textrm{.}</math>}}
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and then cancel the common x2-term,
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We can write this equation as <math>64x = 80</math>, which gives
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<math>24x+20=-40x+100</math>
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We can write this equation as
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<math>\text{64}x=\text{8}0</math>, which gives
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<math>x=\frac{80}{64}=\frac{2^{4}\centerdot 5}{2^{6}}=\frac{5}{2^{2}}=\frac{5}{4}</math>
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{{Displayed math||<math>x = \frac{80}{64} = \frac{2^{4}\cdot 5}{2^{6}} = \frac{5}{2^{2}} = \frac{5}{4}\,\textrm{.}</math>}}
Because we squared our original equation (twice), we have to verify the solution
Because we squared our original equation (twice), we have to verify the solution
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<math>x={5}/{4}\;</math>
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<math>x=5/4</math> in order to be able to rule out that we have a spurious root:
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in order to be able to rule out that we have a false root:
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LHS=
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<math>\begin{align}
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& =\sqrt{\frac{5}{4}+1}+\sqrt{\frac{5}{4}+5}=\sqrt{\frac{9}{4}}+\sqrt{\frac{25}{4}} \\
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& =\frac{3}{2}+\frac{5}{2}=\frac{8}{2}=4= \\
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\end{align}</math>
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=RHS
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{{Displayed math||<math>\begin{align}
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\text{LHS} &= \sqrt{\frac{5}{4}+1} + \sqrt{\frac{5}{4}+5} = \sqrt{\frac{9}{4}} + \sqrt{\frac{25}{4}}\\[10pt]
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&= \frac{3}{2} + \frac{5}{2} = \frac{8}{2} = 4 = \text{RHS}\,\textrm{.}
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\end{align}</math>}}
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Thus, the equation has the solution
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Thus, the equation has the solution <math>x=5/4\,</math>.
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<math>x={5}/{4}\;</math>.
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Version vom 13:17, 1. Okt. 2008

This equation differs from earlier examples in that it contains two root terms, in which case it is not possible to get rid of all square roots at once with one squaring, but rather we need to work in two steps.

Squaring once gives

Vorlage:Displayed math

and expanding the left-hand side gives

Vorlage:Displayed math

which, after simplification, results in the equation

Vorlage:Displayed math

Moving all the terms, other than the root term, over to the right-hand side,

Vorlage:Displayed math

and squaring once again,

Vorlage:Displayed math

at last gives an equation that is completely free of root signs,

Vorlage:Displayed math

Expand both sides

Vorlage:Displayed math

and then cancel the common x²-term,

Vorlage:Displayed math

We can write this equation as \displaystyle 64x = 80, which gives

Vorlage:Displayed math

Because we squared our original equation (twice), we have to verify the solution \displaystyle x=5/4 in order to be able to rule out that we have a spurious root:

Vorlage:Displayed math

Thus, the equation has the solution \displaystyle x=5/4\,.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.2:6“