SOLUTION: Please help me solve this. Thank you! If x=2+sqrt(3) and y= 2-sqrt(3) then solve: x^4+y^4 The answer that i got was 180-12sqrt(3) but i don't think that is correct.

Algebra ->  Real-numbers -> SOLUTION: Please help me solve this. Thank you! If x=2+sqrt(3) and y= 2-sqrt(3) then solve: x^4+y^4 The answer that i got was 180-12sqrt(3) but i don't think that is correct.       Log On


   



Question 982192: Please help me solve this. Thank you!
If x=2+sqrt(3) and y= 2-sqrt(3) then solve:
x^4+y^4
The answer that i got was 180-12sqrt(3) but i don't think that is correct.

Found 3 solutions by josgarithmetic, josmiceli, macston:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
Here is a suggestion, while I hold off from actually showing the work here:

Try using or following the Binomial Theorem to find x^4 and y^4, and then sum them.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+x%5E4+%2B+y%5E4+=+%28+x%5E2+-+y%5E2+%29%5E2+%2B+2x%5E2%2Ay%5E2+
+%28+%28+x+-+y+%29%2A%28+x+%2B+y+%29+%29%5E2+%2B+2x%5E2+%2Ay%5E2+
+%28+2%2Asqrt%283%29%2A4+%29%5E2+%2B+2%2Ax%2Ay%2Ax%2Ay+
+%28+8%2Asqrt%283%29+%29%5E2+%2B+2%2A%28+4-3+%29%2A%28+4-3+%29+
+64%2A3+%2B+2+=+194+
---------------------
check answer:
+x+=+2+%2B+sqrt%283%29+
+x+=+2+%2B+1.732+
+x+=+3.732+
and
+y+=+2+-+sqrt%283%29+
+y+=+2+-+1.732+
+y+=+.2679+
-----------------
+x%5E4+=+3.732%5E4+
+x%5E4+=+193.984+
and
+y%5E4+=+.2679%5E4+
+y%5E4+=+.005151+
+x%5E4+%2B+y%5E4+=+193.984+%2B+.005151+
+x%5E4+%2B+y%5E4+=+193.989+
I think error is due to rounding off

Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
.
x=2%2Bsqrt%283%29
x%5E2=4%2B2sqrt%283%29%2B2sqrt%283%29%2B3
x%5E2=7%2B4sqrt%283%29
x%5E4=%28x%5E2%29%5E2=%287%2B4sqrt%283%29%29%5E2=49%2B28sqrt%282%29%2B28sqrt%283%29%2B48
x%5E4=97%2B56sqrt%283%29
.
Using same method:
y%5E4=97-56sqrt%283%29
So:
X%5E4%2By%5E4=97%2B56sqrt%283%29%2B97-56sqrt%283%29=194