#2(t^(1/5))^2+7t^(1/5)+3=0#? Solve the equation by substitution

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Answer 1 · 2017-12-24T06:29:22

#u=-1/32# or #-243#

In #2(t^(1/5))^2+7t^(1/5)+3-0# substituting #u=t^(1/5)#, the equation becomes

#2u^2+7u+3=0#

or #2u^2+6u+u+3=0#

or #2u(u+3)+1(u+3)=0#

or #(2u+1)(u+3)=0#

i.e. either #u=-1/2# or #u=-3#

If #u=-1/2#, we have #t^(1/5)=-1/2# and #t=(-1/2)^5=-1/32#

and if #u=-3#, #t^(1/5)=-3# or #t=(-3)^5=--243#