SOLUTION: solve y^4-4y^2+4=0(hint: set x=y^2 solve for x, then solve for y after finding x) this makes no sense to me, i feel like it isn't a problem i've learned about

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: solve y^4-4y^2+4=0(hint: set x=y^2 solve for x, then solve for y after finding x) this makes no sense to me, i feel like it isn't a problem i've learned about      Log On


   

Question 187842: solve y^4-4y^2+4=0(hint: set x=y^2 solve for x, then solve for y after finding x)


this makes no sense to me, i feel like it isn't a problem i've learned about
Found 2 solutions by jim_thompson5910, Alan3354:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
y%5E4-4y%5E2%2B4=0 Start with the given equation.


Let x=y%5E2. This means that x%5E2=y%5E4


x%5E2-4x%2B4=0 Make the proper substitutions


Notice we have a quadratic equation in the form of ax%5E2%2Bbx%2Bc where a=1, b=-4, and c=4


Let's use the quadratic formula to solve for x


x+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29 Start with the quadratic formula


x+=+%28-%28-4%29+%2B-+sqrt%28+%28-4%29%5E2-4%281%29%284%29+%29%29%2F%282%281%29%29 Plug in a=1, b=-4, and c=4


x+=+%284+%2B-+sqrt%28+%28-4%29%5E2-4%281%29%284%29+%29%29%2F%282%281%29%29 Negate -4 to get 4.


x+=+%284+%2B-+sqrt%28+16-4%281%29%284%29+%29%29%2F%282%281%29%29 Square -4 to get 16.


x+=+%284+%2B-+sqrt%28+16-16+%29%29%2F%282%281%29%29 Multiply 4%281%29%284%29 to get 16


x+=+%284+%2B-+sqrt%28+0+%29%29%2F%282%281%29%29 Subtract 16 from 16 to get 0


x+=+%284+%2B-+sqrt%28+0+%29%29%2F%282%29 Multiply 2 and 1 to get 2.


x+=+%284+%2B-+0%29%2F%282%29 Take the square root of 0 to get 0.


x+=+%284+%2B+0%29%2F%282%29 or x+=+%284+-+0%29%2F%282%29 Break up the expression.


x+=+%284%29%2F%282%29 or x+=++%284%29%2F%282%29 Combine like terms.


x+=+2 or x+=+2 Simplify.


So the solution is x=2


However, we want to solve for "y"


x=y%5E2 Go back to the second equation.


2=y%5E2 Plug in x=2


y%5E2=2 Rearrange the equation


y=sqrt%282%29 or y=-sqrt%282%29 Take the square root of both sides (don't forget the "plus/minus")


So the solutions are y=sqrt%282%29 or y=-sqrt%282%29

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
solve y^4-4y^2+4=0(hint: set x=y^2 solve for x, then solve for y after finding x)
------------------
Sub x for y^2
x^2 - 4x + 4 = 0
(x-2)*(x-2) = 0
x = 2
------
y^2 = 2
y = +sqrt(2)
y = -sqrt(2)