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Kisachek [45]
3 years ago
8

Does anyone understand exactly what is going on in the example and how to use it for the next two problems? Can anyone explain t

his value substitution method? Thanks!

Mathematics
1 answer:
Sunny_sXe [5.5K]3 years ago
6 0
From the looks of it, it is just smart substitution.  Basically, you want to pick values for "x" in such a way that one of the variables disappear.  Do this enough times and you can get a solvable set of equations.

So for (2)
Pick x = 1 (doing this removes the a variable)
We get:
4b + c = 1

Now pick x = -1 (doing this removes the b variable)
We get: 
4a + c = 1

Now there is no way to remove the c variable so let's just pick x = 0. 
We get:
a + b + c = 0

Now we have 3 set of equations and 3 variables.

Solve this by using Gauss-Jordan elimination or whatever other technique you know (Or just use an online matrix calculator).

You should get:
a = 1/2
b = 1/2
c = -1
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Answer:

1)  \dfrac{2}{\sqrt{5} }  = \dfrac{2 \cdot \sqrt{5} }{5}

2)  -\dfrac{5}{\sqrt{3} } = -\dfrac{5 \cdot \sqrt{3} }{3}

3)  \dfrac{\sqrt{2} + \sqrt{5}  }{\sqrt{10} } =\dfrac{\sqrt{5}  }{5} + \dfrac{ \sqrt{2}  }{2}

4)  \dfrac{3 + \sqrt{2} }{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = \sqrt{3} + \dfrac{\sqrt{6} }{3}

5)  \dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2}  }= \dfrac{\sqrt{15} - \sqrt{6} }{3}

Step-by-step explanation:

The rationalization of the denominator of the surds are found as follows;

1) \dfrac{2}{\sqrt{5} }

\dfrac{2}{\sqrt{5} } \times \dfrac{\sqrt{5} }{\sqrt{5} } = \dfrac{2 \cdot \sqrt{5} }{5}

\dfrac{2}{\sqrt{5} }  = \dfrac{2 \cdot \sqrt{5} }{5}

2) -\dfrac{5}{\sqrt{3} }

-\dfrac{5}{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = -\dfrac{5 \cdot \sqrt{3} }{3}

-\dfrac{5}{\sqrt{3} } = -\dfrac{5 \cdot \sqrt{3} }{3}

3) \dfrac{\sqrt{2} + \sqrt{5}  }{\sqrt{10} }

\dfrac{\sqrt{2} + \sqrt{5}  }{\sqrt{10} } \times \dfrac{ \sqrt{10}  }{\sqrt{10} } = \dfrac{\sqrt{20} + \sqrt{50}  }{10 } = \dfrac{2\cdot \sqrt{5} + 5 \cdot \sqrt{2}  }{10} = \dfrac{\sqrt{5}  }{5} + \dfrac{ \sqrt{2}  }{2}

\dfrac{\sqrt{2} + \sqrt{5}  }{\sqrt{10} } =\dfrac{\sqrt{5}  }{5} + \dfrac{ \sqrt{2}  }{2}

4) \dfrac{3 + \sqrt{2} }{\sqrt{3} }

\dfrac{3 + \sqrt{2} }{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = \dfrac{3 \cdot \sqrt{3}+\sqrt{6}  }{3 } = \sqrt{3} + \dfrac{\sqrt{6} }{3}

\dfrac{3 + \sqrt{2} }{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = \sqrt{3} + \dfrac{\sqrt{6} }{3}

5) \dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2}  }

\dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2}  } = \dfrac{\sqrt{5} - \sqrt{2} }{\sqrt{5} - \sqrt{2} }  = \dfrac{\sqrt{15} -\sqrt{6} }{5 - 2} = \dfrac{\sqrt{15} - \sqrt{6} }{3}

\dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2}  }= \dfrac{\sqrt{15} - \sqrt{6} }{3}

6) \dfrac{\sqrt{7} }{\sqrt{3} - \sqrt{5}  }

\dfrac{\sqrt{7} }{\sqrt{3} - \sqrt{5}  } \times \dfrac{\sqrt{3} + \sqrt{5}}{\sqrt{3} + \sqrt{5}}  = \dfrac{\sqrt{21} + \sqrt{35}}{{3} + {5}} = \dfrac{\sqrt{21} + \sqrt{35}}{8}

\dfrac{\sqrt{7} }{\sqrt{3} - \sqrt{5}  } \times \dfrac{\sqrt{3} + \sqrt{5}}{\sqrt{3} + \sqrt{5}}  =\dfrac{\sqrt{21} + \sqrt{35}}{8}

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