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Proving Two Radical Expressions Are Equivalent

Date: 06/28/2005 at 18:28:00
From: Mike
Subject: equivalent square root expressions

When calculating the radius of a regular icosahedron given the length
of an edge, I calculated an answer involving radicals and to check it
I looked for confirmation on the internet.  The radical answer I found
looked different than mine, but my calculator showed the decimal
versions of the two answers to be the same up until the 14th place.  I
suspect the expressions are equivalent but I cannot algebraically show
it. 
     
                       5-sqrt(5)           5+2*sqrt(5)        
My answer:  .05{ sqrt[ --------- ] + sqrt[ ----------- ] }
                          10                    5

               sqrt[10+2*sqrt(5)]
Their answer:  ------------------
                       4
                                                         
The decimal approximations I got were:

Mine:    0.95105651629518 
Theirs:  0.95105651629515

Can you help me show that the two radical expressions are equal? 
Thanks.  Mike



Date: 06/29/2005 at 21:03:33
From: Doctor Peterson
Subject: Re: equivalent square root expressions

Hi, Mike.

There are a couple ways to prove it; I'll show you the method that can
be used to actually simplify such a sum.  Part of the technique can be
seen here:

  Simplifying Radicals
    http://mathforum.org/library/drmath/view/52660.html 
  
First, let's set the two expressions equal:

           5-sqrt(5)           5+2*sqrt(5)        sqrt[10+2*sqrt(5)]     
  .5{sqrt[ --------- ] + sqrt[ ----------- ]} =?  ------------------
              10                    5                      4

Now, let's see if we can transform the left side into the right.  
First I'll pull out the denominators from the radicals so we can have 
a common denominator:

           5-sqrt(5)           5+2*sqrt(5)     
  .5{sqrt[ --------- ] + sqrt[ ----------- ]}
              10                    5

    sqrt[5 - sqrt(5)]   sqrt[5 + 2 sqrt(5)]
  = ----------------- + -------------------
       2 sqrt(10)            2 sqrt(5)

    sqrt[5 - sqrt(5)] + sqrt[10 + 4 sqrt(5)]
  = ----------------------------------------
                  2 sqrt(10)

(We could simplify this further by dividing top and bottom by sqrt(5),
but I won't bother.)

Now let's simplify the top by squaring it and taking the square root
of the result.  (We have to be sure that it is really the positive
square root that we will get, but that's obvious in this case.)

  (sqrt[5 - sqrt(5)] + sqrt[10 + 4 sqrt(5)])^2

  = (5-sqrt(5)) + 2 sqrt[5-sqrt(5)]sqrt[10+4 sqrt(5)] + (10+4 sqrt(5))

  = 15 + 3 sqrt(5) + 2 sqrt[(5-sqrt(5))(10+4 sqrt(5))]

  = 15 + 3 sqrt(5) + 2 sqrt[30 + 10 sqrt(5)]

We'll be taking the square root of this, but first we need to simplify
it more!  Let's pull out that last complicated radical and apply the
technique from the page I referred to above.

We hope we can simplify it as a sum of radicals:

  sqrt[30 + 10 sqrt(5)] = sqrt(a) + sqrt(b)

Squaring both sides,

  30 + 10 sqrt(5) = a + 2sqrt(a)sqrt(b) + b

so

  a + b = 30
  2sqrt(a)sqrt(b) = 10 sqrt(5)

Squaring the second equation, it becomes

  4ab = 500

or

  ab = 125

Solving the first equation for b and substituting in the second,

  a(30 - a) = 125

  a^2 - 30a + 125 = 0

Solving this for a, we get a=5 or a=25, and thus b=25 or b=5.  This
tells us that

  sqrt[30 + 10 sqrt(5)] = sqrt(25) + sqrt(5) = 5 + sqrt(5)

That's a nice simplification!  Now let's put that back into our 
earlier work:

  (sqrt[5 - sqrt(5)] + sqrt[10 + 4 sqrt(5)])^2

  = 15 + 3 sqrt(5) + 2 sqrt[30 + 10 sqrt(5)]

  = 15 + 3 sqrt(5) + 2 [5 + sqrt(5)]

  = 25 + 5 sqrt(5)

So

  sqrt[5 - sqrt(5)] + sqrt[10 + 4 sqrt(5)]

  = sqrt[25 + 5 sqrt(5)]

Now we can put this into the original expression:

  sqrt[5 - sqrt(5)] + sqrt[10 + 4 sqrt(5)]
  ----------------------------------------
                2 sqrt(10)

    sqrt[25 + 5 sqrt(5)]
  = --------------------
         2 sqrt(10)

    sqrt(5) sqrt[5 + sqrt(5)]
  = -------------------------
         2 sqrt(5)sqrt(2)

    sqrt(2) sqrt[5 + sqrt(5)]
  = -------------------------
               4

    sqrt[10 + 2 sqrt(5)]
  = --------------------
             4

And we're done!


- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 07/01/2005 at 11:08:47
From: Mike
Subject: Thank you (equivalent square root expressions)

Thank you, Dr. Peterson.  I would have never solved the problem by
myself.  I will never cease to be amazed by the beauty of mathematics. 
Your job must be very interesting and rewarding.  Thanks!
Associated Topics:
High School Square & Cube Roots
Middle School Square Roots

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