Question

Question:

Answer 1

\[x ^{\frac{ 1 }{ 2 }}-16x ^{\frac{ 1 }{ 4 }}+55=0\]

Answer 2

Should I square both sides for the first step?

Answer 3

combine the exponents first

Answer 4

how so, if one is 1/2 and the other 1/4

Answer 5

do a change of variable -> x^1/2=y
and solve for y

Answer 6

Do you know how to do quadratic formula?

Answer 7

** x^1/4 = y
then x^1/2=y^2

Answer 8

Is there a different way? it's for a friend in intermediate algebra i doubt they have done this.

Answer 9

or does that get covered in intermediate algebra i forget

Answer 10

y^2-16*y+55=0
(y-11)(y-5)=0

Answer 11

yeah i know how to do that method i'm looking for something "easier"

Answer 12

Yea the way saiberz did it is easier than quadratic. When you find y in their example it will equal x^(1/4) so your answers will need to be put to the 4th power

Answer 13

no quadratic is easier, how does that go, please show me

Answer 14

I mean easier cause quadratic is covered first

Answer 15

Okay, one second.

Answer 16

thank you lasttccasey i appreciate your time and patience :-)

Answer 17

\[x^{1/2}-16x^{1/4}+55=0\] Quadratic Equation \[x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]a=1 b=-16 c=55 Plug these into the equation and you will get 5 and 11 as your two answers for x but x in this case is actually:\[x^{1/4}\]so your solutions will be:\[x^{1/4}=5 ,11\] to get x just put the 5 and 11 to the 4th power so x = 625 and 14,641

Answer 18

Follow all that?

Answer 19

LOL I had a feeling that's how it went i just haven't done this since junior high, i do mostly upper division courses. yes i follow thank you!!!

Answer 20

lol yea I don't think there a much easier way but you're welcome!

Answer 21

okay so this is why we can use the quadratic equation because it can be rewritten:
\[(x^{-1})^2+16(x ^{-1})^4+55=0\]
so wouldn't our solution at the end using the quadratic equation be solving for x^(-1)?

Answer 22

Not exactly,\[(x^{-1})^2 = \frac{1}{x^2}\]Essentially what you are doing is substituting for x^(1/4):
Lets say this..\[y = x^{1/4}\]So this next step just shows all your x's that are raised to the 1/4th power: \[x^{1/2}-16x^{1/4}+55=0 \rightarrow (x^{1/4})^2-16(x^{1/4})^1+55=0\]From here just substitute in for y and you get: \[y = x^{1/4} \rightarrow (y)^2-16(y)^1+55 =0\]From here we use quadratic formula as stated earlier: \[y=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]Notice, the trick is that we are solving for y here which is actually :\[y = x^{1/4}\]
Basically here is your final solution formula:\[x^{1/4} =\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]

Answer 23

A quadratic equation is written as a( )^2 +b( )^1 + c=0
our ( ) was (x^-1)^2=x^-2, however we had a power of 4 so we square again
(x^-2)^2=x^-4 that's what I was looking for thanks anyway, your explanation is nice but I wasn't looking for that.