Solve (sic) #root3x/(root3(36-x^2))# ?

Question

205 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-10T03:59:22

Expression #=root3(x/((6-x)(6+x))#

First a couple of comments.

(i) Since this is not an equation it cannot be "solved". It can, however, be expressed in an (arguably) simpler form.

(ii) I assume the expression to be: #root3x/(root3(36-x^2))#

Expression #= root3x/(root3(36-x^2))#

Remember #rootna/rootnb = rootn(a/b)#

Hence, Expression#= root3(x/(36-x^2))#

Also, #(36-x^2) = (6^2-x^2) = (6-x)(6+x)#

#:.# Expresion #= root3(x/((6-x)(6+x))#