SOLUTION: 4x^4-16x^2+15=0, the answer I got was sqrt6/2, sgrt-6/2, sqrt10/2, sqrt-10/2, is that correct?

Algebra ->  Radicals -> SOLUTION: 4x^4-16x^2+15=0, the answer I got was sqrt6/2, sgrt-6/2, sqrt10/2, sqrt-10/2, is that correct?      Log On


   

Question 158768: 4x^4-16x^2+15=0, the answer I got was sqrt6/2, sgrt-6/2, sqrt10/2, sqrt-10/2, is that correct?
Answer by gonzo(654) About Me  (Show Source):
You can put this solution on YOUR website!
i got x=%2Bsqrt%285%2F2%29 and x=-sqrt%285%2F2%29
i also got x=%2Bsqrt%283%2F2%29 and x=-sqrt%283%2F2%29
here's how.....
4%2Ax%5E4+-+16%2Ax+%2B+15+=+%282x%5E2-5%29%2A%282x%5E2-3%29+=+0
if the whole thing equals 0 then at least one of the factors must = 0.
solving for the first factor = 0, i get
%282%2Ax%5E2-5%29+=+0 becomes %282x%5E2%29=5 becomes x%5E2=%285%2F2%29
and
solving for the second factor = 0, i get
%282%2Ax%5E2-3%29+=+0 becomes %282x%5E2%29=3 becomes x%5E2=%283%2F2%29
plugging x%5E2=%285%2F2%29 into the original equation, i get
4%2A%285%2F2%29%5E2+-+16%2A%285%2F2%29+%2B+15+=+0
this becomes
4%2A%2825%2F4%29+-+80%2F2+%2B+15+=+0
this becomes -15%2B15=0 proving first factor is good.
plugging x%5E2=%283%2F2%29 into the original equation, i get
4%2A%283%2F2%29%5E2+-+16%2A%283%2F2%29+%2B+15+=+0
this becomes
4%2A%289%2F4%29+-+48%2F2+%2B+15+=+0
this becomes -15%2B15=0 proving second factor is good.
if x%5E2+=+%285%2F2%29 and x%5E2+=+%283%2F2%29 are the correct answers (proven above), then
x = +/- sqrt%285%2F2%29 and x = +/- sqrt%283%2F2%29 are the correct values for x.