SOLUTION: Solve {{{ 3t^2 + 2 = 9t^4 }}} My working: {{{ (3t^2 + 1)(3t^2 - 2) = 0 }}} {{{ 3t^2 - 2 = 0 }}} {{{ t^2 = 2/3 }}} {{{ t = - sqrt(2/3) }}} & {{{ t = sqrt(2/3) }}} The

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Solve {{{ 3t^2 + 2 = 9t^4 }}} My working: {{{ (3t^2 + 1)(3t^2 - 2) = 0 }}} {{{ 3t^2 - 2 = 0 }}} {{{ t^2 = 2/3 }}} {{{ t = - sqrt(2/3) }}} & {{{ t = sqrt(2/3) }}} The      Log On


   

Question 742642: Solve +3t%5E2+%2B+2+=+9t%5E4+
My working:
+%283t%5E2+%2B+1%29%283t%5E2+-+2%29+=+0+
+3t%5E2+-+2+=+0+
+t%5E2+=+2%2F3+
+t+=+-+sqrt%282%2F3%29+ & +t+=+sqrt%282%2F3%29+

The textbook answer is +t+=+-+sqrt%286%29%2F2+ & +t+=+sqrt%286%29%2F2+. Please help me find where I went wrong. Thanks.
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
"+3t%5E2+%2B+2+=+9t%5E4+"
"My working:
+%283t%5E2+%2B+1%29%283t%5E2+-+2%29+=+0+"

,..but you do not have a source for the difference of two squares.

3t%5E2+%2B+2+=+9t%5E4
9t%5E4-3t%5E2-2=0
Treating t%5E2 as the variable and resorting to solution to quadratic formula,
you find t%5E2=-1%2F3 or t%5E2=2%2F3.


No REAL solution from t%5E2=-1%2F3, but for the other intermediary result, highlight%28t=-sqrt%282%2F3%29%29 or highlight%28t=sqrt%282%2F3%29%29.

One last detail to finish you may want is to rationalize the denominator.
highlight%28%28sqrt%282%2F3%29%29%2A%28sqrt%283%29%2Fsqrt%283%29%29=sqrt%286%29%2F3%29
and similarly for the negative t result.