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

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Answer 1 · 2017-12-26T07:25:43

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

Let #u=t^(1/5)#

The equation then becomes

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

Using the quadratic formula #x=(-b+-sqrt(b^2-4ac)]/(2a)#

We find that #u=-1/2 or u=-3#

Plugging in #u# for #t#, we get

#t^(1/5)=-1/2# or #t^(1/5)=-3#

With a calculator, you can take the #root(1/5)# of both numbers, and you will have two solutions for #t#:

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