Question

solve for x: x^(1/2) - 5x^(1/4) + 6 = 0

Answer 1

x= 16 and 81

Answer 2

Can you tell me how you did that?

Answer 3

I could, but you wouldnt like it.

Answer 4

My bad you are right

Answer 5

I need to know how you came about the solution, as I don't understand how to do the problem :/

Answer 6

You can use substitution a=x^(1/4) and a^2=x^(1/2)

Answer 7

Let u = x^(1/4)
u^2-5 u+6 = 0
u^2-5 u = -6
u^2-5 u+25/4 = 1/4
(u-5/2)^2 = 1/4
abs(u-5/2) = 1/2
u-5/2 = -1/2 or u-5/2 = 1/2
u = 2 or u-5/2 = 1/2
Replace u = x^(1/4):
x^(1/4) = 2 or u-5/2 = 1/2
Let x = r (cos(theta)+i sin(theta)) and write phasor form:
(r (cos(theta)+i sin(theta)))^(1/4) = 2 (cos(0)+i sin(0))
Use de Moivre's theorem:
r^(1/4) (cos(theta/4)+i sin(theta/4)) = 2 (cos(0)+i sin(0))
Equating moduli and arguments gives:
r^(1/4) = 2 and theta/4 = 2 pi k
Solutions:
r = 16 and theta_1 = 0
Then the equation x^(1/4) = 2 has the solutions:
x = 16 or u-5/2 = 1/2
x = 16 or u = 3
Replace u = x^(1/4):
x = 16 or x^(1/4) = 3
Let x = r (cos(theta)+i sin(theta)) and write 3 in phasor form:
(r (cos(theta)+i sin(theta)))^(1/4) = 3 (cos(0)+i sin(0))
Use de Moivre's theorem:
r^(1/4) (cos(theta/4)+i sin(theta/4)) = 3 (cos(0)+i sin(0))
Equating moduli and arguments gives:
r^(1/4) = 3 and theta/4 = 2 pi k
Solutions:
r = 81 and theta_1 = 0
x = 16 or x = 81

Answer 8

Thats how I did it... like I said, you probably wouldnt like it, but it's the quickest method (that i see right now)

Answer 9

Yeah I see how that works. I just wonder if there is a different way to do it as this is for an Intermediate Algebra class and cosines and sines aren't used at all.

Answer 10

you can try and combine the fractional powers and then evaluate only on the positives and then write a proof that would allow you to operate it on the negative side too, but that seems like a lot more work to me O_o

Answer 11

Hmm this just doesn't seem to fit in with what is being taught in my class at all.

Answer 12

Well, most algebra classes/books/teachers are stupid and don't realize that :
\[\sqrt {x^2} \ne x\]
So they do things that I cant, because I know that's the case O_o

Answer 13

rather than all that sine cosine crap I just put u=3 and u=2 then replaced u back with x^(1/4) then I just did 3^4 and 2^4 to get the same answers

Answer 14

@Docswoops don't worry. agreene just try to impress you with all his knowledge :-))) You already had an answer.

Answer 15

In any case I understand how to come to the answer now lol.

Answer 16

@MarinaDL he is trying to find an answer he understands how to get... and I'm trying to think of a solution for him. Don't presume you know what I'm doing at any given time.

@sclowers
You're probably on the right tract but your Lemma has the incorrect powers :X

Answer 17

Calm down guys it all makes sense now haha! Honestly I just didn't notice replacing x^(1/4) with "u".... was simple after that.

Answer 18

Thats what I did, but I had to use de Moivres because of the above situation with roots not cancelling for the negative numbers. (else you are needing a proof it hold for the negatives)

Answer 19

@agreene Thanks for detailed explanation.

Answer 20

But, alas, if you understand it, then meh, more power to yah! lol... I'm probably being a douche right now.

Answer 21

I'm assuming that level of answer is above my class level.