SOLUTION: How would I solve this quadratic equation?: {{{y^3-28y^1.5+27=0}}} (I used decimal exponents because i have no idea how to write fractional exponents on a computer) I have

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: How would I solve this quadratic equation?: {{{y^3-28y^1.5+27=0}}} (I used decimal exponents because i have no idea how to write fractional exponents on a computer) I have      Log On


   

Question 120103: How would I solve this quadratic equation?:
y%5E3-28y%5E1.5%2B27=0
(I used decimal exponents because i have no idea how to write fractional exponents on a computer)
I have a very rough idea of where to start but am still having a lot of trouble, please help!
James


Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for y:
y%5E3-28y%5E%281.5%29%2B27+=+0
You can solve this problem by substituting another varible for y as follows:
Let z+=+y%5E%281.5%29, then z%5E2+=+y%5E3
z%5E2-28z%2B27+=+0 Now you can solve this quadratic equation by factoring:
%28z-1%29%28z-27%29+=+0 Apply the zero product principle:
z-1+=+0 or z-27+=+0
If z-1+=+0 then z+=+1
If z-27+=+0 then z=27
But we said that z+=+y%5E1.5, so we make the substitution back:
y%5E1.5+=+1 or y%5E1.5+=+27
Let's rewrite the exponents:
y%5E1.5+=+y%5E%283%2F2%29
y%5E%283%2F2%29+=+1 which can also be written:
%28sqrt%28y%29%29%5E3+=+1 Take the cube-root of both sides.
sqrt%28y%29+=+1 Now square both sides.
y+=+1 and...
y%5E%283%2F2%29+=+27
%28sqrt%28y%29%29%5E3+=+27 Take the cube-root of both sides.
sqrt%28y%29+=+3 Now square both sides.
y+=+9
The solutions are:
y+=+1, y+=+9