Question 162013
1)


For these type of problems, it is <b>crucial</b> to remember that {{{sqrt(x^2)=x}}} where {{{x>=0}}} (ie it's not negative)


{{{sqrt(16x^2y^4)}}} Start with the given expression



{{{sqrt(4^2*x^2y^4)}}} Rewrite {{{16}}} as {{{4^2}}}



{{{sqrt(4^2*x^2(y^2)^2)}}} Rewrite {{{y^4}}} as {{{(y^2)^2}}}



{{{sqrt(4^2)*sqrt(x^2)*sqrt((y^2)^2)}}} Break up the square root.



{{{4xy^2}}} Use the formula above to simplify each square root. So in this case {{{sqrt(4^2)=4}}}, {{{sqrt(x^2)=x}}}, and {{{sqrt((y^2)^2)=y^2}}}



So {{{sqrt(16x^2y^4)=4xy^2}}} every variable is nonnegative




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2)


{{{sqrt(12)-sqrt(18)+3*sqrt(50)+sqrt(75)}}} Start with the given expression


Now we must factor each number in the square roots into two number where one MUST be a perfect square.


{{{sqrt(4*3)-sqrt(9*2)+3*sqrt(25*2)+sqrt(25*3)}}} Factor {{{12}}} into {{{4*3}}}. Factor {{{18}}} into {{{9*2}}}. Factor {{{50}}} into {{{25*2}}}. Factor {{{75}}} into {{{25*3}}}. 



{{{sqrt(4)*sqrt(3)-sqrt(9)*sqrt(2)+3*sqrt(25)*sqrt(2)+sqrt(25)*sqrt(3)}}} Break p the square roots



{{{sqrt(2^2)*sqrt(3)-sqrt(3^2)*sqrt(2)+3*sqrt(5^2)*sqrt(2)+sqrt(5^2)*sqrt(3)}}} Rewrite {{{4}}} as {{{2^2}}}. Rewrite {{{9}}} as {{{3^2}}}. Rewrite {{{25}}} as {{{5^2}}}


{{{2*sqrt(3)-3*sqrt(2)+3*5*sqrt(2)+5*sqrt(3)}}} Take the square root of the perfect squares to simplify the square roots. So {{{sqrt(2^2)=2}}}, {{{sqrt(3^2)=3}}}, and {{{sqrt(5^2)=5}}}



{{{2*sqrt(3)-3*sqrt(2)+15*sqrt(2)+5*sqrt(3)}}} Multiply



{{{(-3*sqrt(2)+15*sqrt(2))+(2*sqrt(3)+5*sqrt(3))}}} Group the common terms.



{{{12*sqrt(2)+7*sqrt(3)}}} Combine the common terms.



So {{{sqrt(12)-sqrt(18)+3*sqrt(50)+sqrt(75)=12*sqrt(2)+7*sqrt(3)}}}