How do you evaluate 2√44x^3 -√7 - √99x^3 +√63?

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Answer 1 · 2015-11-08T15:46:24

#x sqrt(11x)+2sqrt7)#

#44 = 4 times 11 " and 4 is " 2^2#
#99 = 9 times 11 " and 9 is " 3^2#

Both 4 and 9 have square roots

#63 = 3 times 21 = 3 times 3 times 7 =3^2 times 7#

#2 sqrt(44x^3) - sqrt(7) - sqrt(99x^3) + sqrt(63)" "# becomes

#2sqrt(2^2 times 11x^3) -sqrt(7)-sqrt(3^2 times 11x^3) +sqrt(3^2 times 7)#

#4sqrt(11x^3) -3sqrt(11x^3)+3sqrt(7)-sqrt(7)#

#sqrt(11x^3)+2sqrt(7)#

But #x^3 = x^2x# giving:

#x sqrt(11x)+2sqrt(7)#

There are no more squares that can be factored out of 11 and 7 so this is the final answer/