Infinite square roots
To 2 decimal places, approximate
3 + 3 + 3 + 3 + ⋯ .
This problem is part of the set Hard Equations .
The answer is 2.30.
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6 solutions
what do you do if it's a cube root instead
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Nothing but to solve the equation:x³-x-3＝0,where x＞0(cuz the number in the cube is a positive number),then you can figure out x≈1.671699882
Let x = 3 + 3 + 3 + 3 + . . . . .
Then x 2 = 3 + 3 + 3 + 3 + . . . . . which is x 2 − x − 3 = 0
Solving this quadratic equation gives − 1 . 3 and 2 . 3
Of course x has to be positive so the answer is 2 . 3
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I basically produced the same solution :). Isn't funny how -0.25 will produce a real value, yet − 1 / 4 = 2 i , is imaginary ?
Let the given expression be X
Then
X^2 = 3 + X Then X = 2.3027756
√(3+√3+...)=x x²=3+x x²-x-3=0 (1+√13)/2 approximately 2.30
Set the expression equal to x. x = (x+3)^.5, because the expression is infinite. Square both sides and rearrange in quadratic form: x^2 - x - 3. The only positive root of the quadratic is approximately 2.30 to 2 decimal places.
Well let this be x. So we get 3+x =x² (after squaring) X²-x-3=0 By -b±√D/2a So it would be 1±√13/2 Ie. 1+3.60/2 =4.60/2 =2.30