Question 895379
{{{ sqrt(5) }}} has a value between
{{{ 2 }}} and {{{ 3 }}}, but is less than {{{ 3}}}
so, {{{ 3 - sqrt(5) }}} is positive, and the 
absolute value is not needed
{{{ abs( 3 - sqrt(5) ) + 4 = 3 - sqrt(5) + 4 }}}
{{{ 3 - sqrt(5) + 4  = 7 - sqrt(5) }}}
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{{{ sqrt(10) }}} has a value between {{{ 3 }}}
and {{{ 4 }}} and is greater than {{{ 3 }}}, so
{{{ abs( 3 - sqrt(10) ) = sqrt(10) - 3 }}}, so
{{{ abs( 3 - sqrt(10)) + 4 = sqrt(10) - 3 + 4 }}}
{{{  sqrt(10) - 3 + 4 = sqrt(10) + 1 }}}