Question 186806
<font size = 8 color = "red"><b>Solution by Edwin:</b></font>
<pre><font size = 4 color = "indigo"><b>
{{{sqrt(128) - sqrt(75) - sqrt(98)}}}

Since this involves square roots, we make a list of squares
greater than {{{1^2}}}.

{{{2^2 = 4}}}
{{{3^2=9}}}
{{{4^2=16}}}
{{{5^2=25}}}
{{{6^2=36}}}
{{{7^2=49}}}
{{{8^2=64}}}
{{{9^2=81}}}
{{{10^2=100}}}
{{{11^2=121}}}
{{{12^2=144}}}

You have 128, 75, and 98

For 128, find the largest square on the right 
in the list above that will divide evenly into
128.  This is 64.  So we write 128 as 64*2.

For 75, find the largest square on the right 
in the list above that will divide evenly into
75.  This is 25.  So we write 75 as 25*3.

For 98, find the largest square on the right 
in the list above that will divide evenly into
98.  This is 49.  So we write 98 as 49*2.

So instead of

{{{sqrt(128) - sqrt(75) - sqrt(98)}}}

we write

{{{sqrt(64*2) - sqrt(25*3) - sqrt(49*2)}}}

Now we write the square roots of the products as
the product of the square roots:

{{{sqrt(64)sqrt(2) - sqrt(25)sqrt(3) - sqrt(49)sqrt(2)}}}

We know the square roots of 64,25, and 49 are 8,5, and 7,
so now we have

{{{8sqrt(2) - 5sqrt(3) - 7sqrt(2)}}}

Now the first and third terms are like terms,
so 

8 square roots of 2 minus 7 square roots of two equals
1 square root of 2.  So combining the first and third
terms we have 

{{{1sqrt(2) - 5sqrt(3)}}}

but we don't need the 1

{{{sqrt(2) - 5sqrt(3)}}}

You cannot go further than this, so you stop here!
This is the simplest radical form.

Edwin</pre>