How do you solve #x^(3/2) -2x^(3/4) +1 = 0#?

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Answer 1 · 2017-01-25T19:52:14

#x=1#

Let #t = x^(3/4)#

Then:

#x^(3/2) = x^(3/4*2) = (x^(3/4))^2 = t^2#

and our equation becomes:

#0 = t^2-2t+1 = (t-1)^2#

This has one (repeated) root, namely #t=1#

So:

#x^(3/4) = 1#

If #x >= 0# then:

#x = x^1 = x^(3/4*4/3) = (x^(3/4))^(4/3) = 1^(4/3) = 1#

This is the only Real root.