How do you simplifiy #\sqrt { 1573x ^ { 7} y ^ { 2} z ^ { 4} }#?

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Answer 1 · 2018-04-09T23:09:47

#11|\x^3y|z^2sqrt(13x)#

Let's find the perfect squares in the radical. Recall that #sqrt(ab)=sqrt(a*b)#, #sqrt(a^(b+c))=sqrt(a^b*a^c)#, and #(a^b)^c=a^(b*c)=a^(bc)#

#sqrt(121*13*x^6*x*y^2*z^4) rarr# 121, #x^6#, #y^2#, and #z^4# are all perfect squares. They can be taken out of the radical.

#11^2=121#

#(x^3)^2=x^6#

#y^2=y^2#

#(z^2)^2=z^4#

#11|\x^3y|z^2sqrt(13x) rarr# Absolute value bars need to be placed around the odd exponents so that they are positive (not around the even ones)

Answer 2 · 2018-04-10T00:29:59

#sqrt(1573x^7y^2z^4)=11sqrt(13)x^(7/2)|y|z^2#

#sqrt(1573x^7y^2z^4)=sqrt(13)(11^2x^7y^2z^4)^(1/2)#

#=11sqrt(13)x^(7/2)|y|z^2#.