Prove that (17+12√2)^(1/4) - (17-12√2)^(1/4) = 2 ?

Question

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.

Answer 1

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. :)

Answer 2

√√(17+12√2) - √√(17-12√2) =
√√(9 + 2x3x√8 + 8) - √√(9 - 2x3x√8 + 8) =
√√(3 + √8)² - √√(3 - √8)² =

√(3 + 2√2) - √(3 - 2√2) =
√(2 + 2x√2x1 + 1) - √(2 - 2x√2x1 +1) =
√(√2 + 1)² - √(√2 +1)² =

(√2 + 1) - (√2 +1) =
1 + 1 =
2

Answer 3

Let a=17 and b=12√2 and c=(17+12√2)^(1/4) - (17-12√2)^(1/4)

c^2=√(a+b)+√(a-b)-2(a^2-b^2)^(1/4)=√(a+b)+√(a-b)-2

c^2+2=√(a+b)+√(a-b)

c^4=(a+b)+(a-b)+4+2√(a^2-b^2)-4√(a+b)-4√(a-b) =2a+4+2-4*[√(a+b)+√(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+12√2)^(1/4) - (17-12√2)^(1/4)=2

Answer 4

(17+12√2)^(1/4) - (17-12√2)^(1/4) = 2
(17+12(1∙414 213 562...))^(1/4) - (17-12(1∙414 213 562...))^(1/4) = 2
(17+16∙970 562 75)^(1/4) - (17-16∙970 562 715)^(1/4) = 2
(33∙970 562 75)^(1/4) - (0∙029 437 251)^(1/4) = 2
(2∙4142......) - (0∙41421....) = 2
2 = 2

Answer 5

I think that if you multiply both sides by (17+12 sqrt(2))^1/4, wondereful things might happen.

Answer 6

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