Question

Find the value of 12 to the power of (-1+sqrt(5))/2

\Big( \dfrac{-1+\sqrt{5} }{2} \Big) ^{12}

Collected in the board: Square Root

Steven Zheng posted 17 hours ago

Answer 1

Let

a = \dfrac{-1+\sqrt{5} }{2}
(1)

Rearrange the equation

2a+1 = \sqrt{5}

Square both sides

4a^2+4a+1=5

Then,

a^2=1-a
(2)

Square both sides

a^4 = a^2-2a+1
(3)

Substituting (2) to (3) gives

a^4= 2-3a
(4)

Multiplying (2) by (4) gives

a^6 = (1-a)(2-3a)

Expand the right hand side

a^6 = 2-5a+3a^2
(5)

Substituting (2) to (5) gives

a^6 = 2-5a+3-3a = 5-8a
(6)

Squaring both sides of equation (6) gives

a^{12} = 25-80a+64a^2
(7)

Substitute (2) to (7)

a^{12} = 25-80a+64(1-a) = 89-144a
(8)

Substitute the value of a to (8), we get final result

a^{12} = 89 - 144\cdot \dfrac{-1+\sqrt{5} }{2} = 161 - 72\sqrt{5}

that is, 12 to the power of (-1+sqrt(5))/2 is euqal to 161 - 72\sqrt{5}

Steven Zheng posted 17 hours ago

Answer 2

Let

a = \dfrac{-1+\sqrt{5} }{2}
(1)

Then,

\dfrac{1}{a} = \dfrac{1}{ \dfrac{-1+\sqrt{5} }{2}} = \dfrac{1+\sqrt{5} }{2}
(2)

Addition of a and \dfrac{1}{a} gives a Nike equation

a+\dfrac{1}{a} = \sqrt{5}
(3)

Square both sides and move the constant term to the right

a^2+\dfrac{1}{a^2} = 3
(4)

Square both sides and move the constant term to the right

a^4+\dfrac{1}{a^4} =7
(5)

Multiplying (2) by (3) gives

a^6+a^2+\dfrac{1}{a^2} +\dfrac{1}{a^6} = 21
(6)

Substitution (2) to (4) gives

a^6+\dfrac{1}{a^6} = 18
(7)

Squaring both sides of (5) and move the constant term to the right

a^{12}+\dfrac{1}{a^{12}} = 18^2-2=322
(8)

Let

m =a^{12}
(9)

Then a quadratic equation is obtained

m^2-322m+1 = 0

Using the root formula for a quadratic equation

m = \dfrac{322\pm\sqrt{322^2-4} }{2}

Since \dfrac{-1+\sqrt{5} }{2} <1, cancel the root larger than 1. Then we obtained the final result

m = 161-72\sqrt{5}


Steven Zheng posted 17 hours ago

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