SOLUTION: I attempted this problem but I am not sure of the answer. Can someone check to see if my answer is correct. Thanks. 36y^4+36y^3+6y-1= 6y^2+6y-1

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: I attempted this problem but I am not sure of the answer. Can someone check to see if my answer is correct. Thanks. 36y^4+36y^3+6y-1= 6y^2+6y-1      Log On


   

Question 265831: I attempted this problem but I am not sure of the answer. Can someone check to see if my answer is correct. Thanks.
36y^4+36y^3+6y-1= 6y^2+6y-1
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
36y%5E4%2B36y%5E3%2B6y-1=+6y%5E2%2B6y-1

Get 0 on the right:

36y%5E4%2B36y%5E3-6y%5E2=0

factor out 6y%5E2 on the left

6y%5E2%286y%5E2%2B6y-1%29=0

Set the first factor =0

6y%5E2=0
y%5E2=0
y=0, that's one solution.

Set the second factor = 0

6y%5E2%2B6y-1=0

That trinomial will not factor. So we
use the quadratic formula:

y+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ 

y+=+%28-%286%29+%2B-+sqrt%28+%286%29%5E2-4%2A%286%29%2A%28-1%29+%29%29%2F%282%2A%286%29%29+ 

y+=+%28-6+%2B-+sqrt%28+36%2B24%29%29%2F12+

y+=+%28-6+%2B-+sqrt%2860%29%29%2F12+

y+=+%28-6+%2B-+sqrt%284%2A15%29%29%2F12+

y+=+%28-6+%2B-+2%2Asqrt%2815%29%29%2F12+

y+=+%282%28-3+%2B-+sqrt%2815%29%29%29%2F12+

y+=+%28cross%282%29%28-3+%2B-+sqrt%2815%29%29%29%2Fcross%2812%29%5B6%5D+

y=%28-3+%2B-+sqrt%2815%29%29%2F6

So there are three solutions:

y=0, y=%28-3+%2B+sqrt%2815%29%29%2F6, and y=%28-3+-+sqrt%2815%29%29%2F6

Edwin