Problem: solve for x :
x^1/2 +x^1/4 = 20
x^1/2 + x^1/4 = 20
Let's re-write that. We are dealing with
a number, n, which, when squared, equals x.
We have another number, m, which, when squared,
equals n.
n * n = x ... n = x^1/2
m * m = n ... n = m^2
m * m * m * m = x .... m = x^1/4
n + m = 20
m^2 + m = 20
m^2 + m - 20 = 0
(m + 5)(m - 4) = 0
One, or both, of those factors must equal zero.
m + 5 = 0
m = -5
m - 4 = 0
m = 4
If m is -5, m^2 is 25; that's already greater than 20,
so it won't work.
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Correction, added the following day. I was too quick
in rejecting -5 as an answer. Re-visit this equation:
m^2 + m = 20
-5^2 + (-5) = 20
25 - 5 = 20
It actually conforms to the problem requirement.
I am adding a solution at the end of this answer,
based on -5 as a valid factor.
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If m is 4, m^2 is 16. Add m to that, and we have 20.
The original equation is:
n + m = 20
16 + 4 = 20
We saw above that x = n^2.
x = 16^2
x = 256
Also, x = m^4
x = 4^4
x = 256
x^1/2 + x^1/4 = 20
256^1/2 + 256^1/4 = 20
* * * * * * * * *
Add that
x = 25^2
x = 625
and
x = -5^4
x = 625
* * * * * * * * *
Answer: x = 256 AND.... x = 625