How do you evaluate square root of #a^(11) + sqrt(a^5)#?

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Answer 1 · 2017-07-06T13:46:37

#a^(5/2)(a^(17/2)+1)# This may also be written as: #sqrt(a^5)(sqrt(a^(17))+1)#

The use of the word 'evaluate' in my book states; 'give value to'. This is not possible unless you know the value of #a#

#color(blue)("Some facts about indices and roots ")#

Consider the example #sqrt(b)# this may also be written as #b^(1/2)#

#root(3)(b)=b^(1/3)#
#root(4)(b)=b^(1/4)#
#root(5)(b)=b^(1/5)#
So the root is the denominator of the 'power'

#sqrt(b^2)=b^(2/2)=b^1=b#
#sqrt(b^3)=b^(3/2)#
#sqrt(b^4)=b^(4/2)=b^2#
and so on

Consider the example #3^3xx3" " =" " 3xx3xx3xx3" "=" "3^4#

3 may, if you so wish, be written as #3^1#

So #3^2xx3" "=3^2xx3^1" "=" "3^(3+1)" "=" "3^4#

So #b^5xxb^7=b^(5+7)=b^(12)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Answering the question")#

Given:#" " a^11+sqrt(a^5)#

Write as:

#[a^11]+[sqrt(x^5)xx1]#

But #a^11# is the same as #a^(22/2)" "=" "a^(5/2+17/2)" "=" "a^(5/2)xxa^(17/2)#

Also #sqrt(a^5)# is the same as #a^(5/2)#

Putting it all together we have

#[a^(5/2)xxa^(17/2)]+[a^(5/2)xx1]#

Factor out the #a^(5/2)# giving:

#a^(5/2)(a^(17/2)+1)#

This may also be written as:

#sqrt(a^5)(sqrt(a^(17))+1)#

NOT CONVINCED THIS HELPS