Lösung 3.2:6

Aus Online Mathematik Brückenkurs 1

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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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{{Abgesetzte Formel||<math>\bigl(\sqrt{x+1}+\sqrt{x+5}\,\bigr)^2 = 4^2</math>}}
and expanding the left-hand side gives
and expanding the left-hand side 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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{{Abgesetzte Formel||<math>\bigl(\sqrt{x+1}\bigr)^2 + 2\sqrt{x+1}\sqrt{x+5} + \bigl(\sqrt{x+5}\bigr)^2 = 16</math>}}
which, after simplification, results in the equation
which, after simplification, results in the equation
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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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{{Abgesetzte Formel||<math>x+1+2\sqrt{x+1}\sqrt{x+5}+x+5=16\,\textrm{.}</math>}}
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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{{Abgesetzte Formel||<math>2\sqrt{x+1}\sqrt{x+5}=-2x+10</math>}}
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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{{Abgesetzte Formel||<math>\bigl(2\sqrt{x+1}\sqrt{x+5}\bigr)^2 = (-2x+10)^2</math>}}
at last gives an equation that is completely free of root signs,
at last gives an equation that is completely free of root signs,
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{{Displayed math||<math>4(x+1)(x+5) = (-2x+10)^{2}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<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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{{Abgesetzte Formel||<math>4(x^{2}+6x+5) = 4x^2-40x+100</math>}}
and then cancel the common ''x''²-term,
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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{{Abgesetzte Formel||<math>24x+20=-40x+100\,\textrm{.}</math>}}
We can write this equation as <math>64x = 80</math>, which gives
We can write this equation as <math>64x = 80</math>, which gives
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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>}}
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{{Abgesetzte Formel||<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
<math>x=5/4</math> in order to be able to rule out that we have a spurious root:
<math>x=5/4</math> in order to be able to rule out that we have a spurious root:
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\text{LHS} &= \sqrt{\frac{5}{4}+1} + \sqrt{\frac{5}{4}+5} = \sqrt{\frac{9}{4}} + \sqrt{\frac{25}{4}}\\[10pt]
\text{LHS} &= \sqrt{\frac{5}{4}+1} + \sqrt{\frac{5}{4}+5} = \sqrt{\frac{9}{4}} + \sqrt{\frac{25}{4}}\\[10pt]
&= \frac{3}{2} + \frac{5}{2} = \frac{8}{2} = 4 = \text{RHS}\,\textrm{.}
&= \frac{3}{2} + \frac{5}{2} = \frac{8}{2} = 4 = \text{RHS}\,\textrm{.}

Version vom 08:41, 22. 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

\displaystyle \bigl(\sqrt{x+1}+\sqrt{x+5}\,\bigr)^2 = 4^2

and expanding the left-hand side gives

\displaystyle \bigl(\sqrt{x+1}\bigr)^2 + 2\sqrt{x+1}\sqrt{x+5} + \bigl(\sqrt{x+5}\bigr)^2 = 16

which, after simplification, results in the equation

\displaystyle x+1+2\sqrt{x+1}\sqrt{x+5}+x+5=16\,\textrm{.}

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

\displaystyle 2\sqrt{x+1}\sqrt{x+5}=-2x+10

and squaring once again,

\displaystyle \bigl(2\sqrt{x+1}\sqrt{x+5}\bigr)^2 = (-2x+10)^2

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

\displaystyle 4(x+1)(x+5) = (-2x+10)^{2}\,\textrm{.}

Expand both sides

\displaystyle 4(x^{2}+6x+5) = 4x^2-40x+100

and then cancel the common x²-term,

\displaystyle 24x+20=-40x+100\,\textrm{.}

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

\displaystyle x = \frac{80}{64} = \frac{2^{4}\cdot 5}{2^{6}} = \frac{5}{2^{2}} = \frac{5}{4}\,\textrm{.}

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:

\displaystyle \begin{align}

\text{LHS} &= \sqrt{\frac{5}{4}+1} + \sqrt{\frac{5}{4}+5} = \sqrt{\frac{9}{4}} + \sqrt{\frac{25}{4}}\\[10pt] &= \frac{3}{2} + \frac{5}{2} = \frac{8}{2} = 4 = \text{RHS}\,\textrm{.} \end{align}

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

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