How do you solve #root4(z^2+6z)=root4(3z+18)#?

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Answer 1 · 2016-09-12T23:26:37

#(root(4)(z^2 + 6z))^4 = (root(4)(3z + 18))^4#

#z^2 + 6z = 3z + 18#

#z^2 + 3z - 18 = 0#

#(z + 6)(z - 3) = 0#

#z = -6 and 3#

Checking, both solutions work, so your solution set is #{-6, 3}#.

Hopefully this helps!