I don't know if I'm getting loved or hated for these kinds of puzzles? Keep up the good work, folks! This can't be any worse than Sudoku.
No way! (Is this response too vague?)
OK, I am wise to these puzzles now....
(1 + √2)^4 = 17 + 12√2
(1 - √2)^4 = 17 - 12√2
Therefore your equation can be rewritten as
(1 + √2)^4^1/4 + (1 - √2)^4^1/4 =
(1 + √2) + (1 - √2) =
2
Hmm, interesting quirk about this puzzle: You can get 2 as a result if you make it an addition as well as a subtraction, since the fourth root can be positive or negative. I copied down the sign wrong, but it gives the same answer. :)
ââ(17+12â2) - ââ(17-12â2) =
ââ(9 + 2x3xâ8 + 8) - ââ(9 - 2x3xâ8 + 8) =
ââ(3 + â8)² - ââ(3 - â8)² =
â(3 + 2â2) - â(3 - 2â2) =
â(2 + 2xâ2x1 + 1) - â(2 - 2xâ2x1 +1) =
â(â2 + 1)² - â(â2 +1)² =
(â2 + 1) - (â2 +1) =
1 + 1 =
Let a=17 and b=12â2 and c=(17+12â2)^(1/4) - (17-12â2)^(1/4)
c^2=â(a+b)+â(a-b)-2(a^2-b^2)^(1/4)=â(a+b)+â(a-b)-2
c^2+2=â(a+b)+â(a-b)
c^4=(a+b)+(a-b)+4+2â(a^2-b^2)-4â(a+b)-4â(a-b) =2a+4+2-4*[â(a+b)+â(a-b)]=2*17 + 4 + 2- 4*[c^2+2] =40-4c^2-8 =32-4c^2
c^4+4c^2-32=0
(c^2-4)(c^2+8)=0
The real solution is c=2
(17+12â2)^(1/4) - (17-12â2)^(1/4)=2
(17+12â2)^(1/4) - (17-12â2)^(1/4) = 2
(17+12(1â414 213 562...))^(1/4) - (17-12(1â414 213 562...))^(1/4) = 2
(17+16â970 562 75)^(1/4) - (17-16â970 562 715)^(1/4) = 2
(33â970 562 75)^(1/4) - (0â029 437 251)^(1/4) = 2
(2â4142......) - (0â41421....) = 2
2 = 2
I think that if you multiply both sides by (17+12 sqrt(2))^1/4, wondereful things might happen.
hi
(V2 +/- 1)^2 = 3 +/- 2V2
(V2 +/- 1)^4 = (3 +/- 2V2)^2 = 17 +/- 12V2
then
(17+12V2)^(1/4) = V2 + 1
(17-12V2)^(1/4) = V2 - 1
difference = 2 ; sum = 2V2
bye
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Answers & Comments
Verified answer
No way! (Is this response too vague?)
OK, I am wise to these puzzles now....
(1 + √2)^4 = 17 + 12√2
(1 - √2)^4 = 17 - 12√2
Therefore your equation can be rewritten as
(1 + √2)^4^1/4 + (1 - √2)^4^1/4 =
(1 + √2) + (1 - √2) =
2
Hmm, interesting quirk about this puzzle: You can get 2 as a result if you make it an addition as well as a subtraction, since the fourth root can be positive or negative. I copied down the sign wrong, but it gives the same answer. :)
ââ(17+12â2) - ââ(17-12â2) =
ââ(9 + 2x3xâ8 + 8) - ââ(9 - 2x3xâ8 + 8) =
ââ(3 + â8)² - ââ(3 - â8)² =
â(3 + 2â2) - â(3 - 2â2) =
â(2 + 2xâ2x1 + 1) - â(2 - 2xâ2x1 +1) =
â(â2 + 1)² - â(â2 +1)² =
(â2 + 1) - (â2 +1) =
1 + 1 =
2
Let a=17 and b=12â2 and c=(17+12â2)^(1/4) - (17-12â2)^(1/4)
c^2=â(a+b)+â(a-b)-2(a^2-b^2)^(1/4)=â(a+b)+â(a-b)-2
c^2+2=â(a+b)+â(a-b)
c^4=(a+b)+(a-b)+4+2â(a^2-b^2)-4â(a+b)-4â(a-b) =2a+4+2-4*[â(a+b)+â(a-b)]=2*17 + 4 + 2- 4*[c^2+2] =40-4c^2-8 =32-4c^2
c^4+4c^2-32=0
(c^2-4)(c^2+8)=0
The real solution is c=2
(17+12â2)^(1/4) - (17-12â2)^(1/4)=2
(17+12â2)^(1/4) - (17-12â2)^(1/4) = 2
(17+12(1â414 213 562...))^(1/4) - (17-12(1â414 213 562...))^(1/4) = 2
(17+16â970 562 75)^(1/4) - (17-16â970 562 715)^(1/4) = 2
(33â970 562 75)^(1/4) - (0â029 437 251)^(1/4) = 2
(2â4142......) - (0â41421....) = 2
2 = 2
I think that if you multiply both sides by (17+12 sqrt(2))^1/4, wondereful things might happen.
hi
(V2 +/- 1)^2 = 3 +/- 2V2
(V2 +/- 1)^4 = (3 +/- 2V2)^2 = 17 +/- 12V2
then
(17+12V2)^(1/4) = V2 + 1
(17-12V2)^(1/4) = V2 - 1
difference = 2 ; sum = 2V2
bye