﻿ Find the value of \sqrt{77+20\sqrt{13} } +\sqrt{77-20\sqrt{13} }

#### Question

Find the value of

\sqrt{77+20\sqrt{13} } +\sqrt{77-20\sqrt{13} }

Collected in the board: Cube root

Steven Zheng posted 5 months ago

Let

a = \sqrt{77+20\sqrt{13} }
(1)
b = \sqrt{77-20\sqrt{13} }
(2)

Addition of the equations after raising a and b to the 3rd power gives

a^3+b^3 = 154
(3)

multiplying (1) with (2) gives

ab = \sqrt{77^2-(20\sqrt{13} )^2} =\sqrt{729} = 9
(4)

Transform the identity of perfect square of binomial to

(a+b)^3 = a^3+b^3+3ab(a+b)
(5)

Let

x = a+b
(6)

Substituting (3), (4) and (6) into (5) results in a depressed cubic equation

x^3-27x-154 = 0
(7)

By observation, x = 7 is one of the solution for equation (7)

Then, divide the polynomial x^3-27x-154 by x-7 to find another factor

(x-7)(x^2+7x+22) = 0
(8)

Since x^2+7x+22 = (x+\dfrac{7}{2})^2 + \dfrac{39}{4} >0 , there is no real solution for the equation x^2+7x+22 = 0

Therefore, there's only one solution x = 7. Now we conclude

\sqrt{77+20\sqrt{13} } +\sqrt{77-20\sqrt{13} } = 7

Steven Zheng posted 5 months ago

Transform the square root binomial to the form of cube of binomial

77+20\sqrt{13} = \Big( \dfrac{7+\sqrt{13} }{2} \Big) ^3

77-20\sqrt{13} = \Big( \dfrac{7-\sqrt{13} }{2} \Big) ^3

Therefore,

\sqrt{77+20\sqrt{13} } +\sqrt{77-20\sqrt{13} } = 7

Steven Zheng posted 5 months ago

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