﻿ Find the value of \sqrt{7+5\sqrt{2}}+\sqrt{7-5\sqrt{2}}

#### Question

Find the value of \sqrt{7+5\sqrt{2}}+\sqrt{7-5\sqrt{2}}

Collected in the board: Cube root

Steven Zheng posted 5 months ago

Let

a =\sqrt{7+5\sqrt{2} }
(1)
b = \sqrt{7-5\sqrt{2} }
(2)

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

a^3+b^3 = 14
(3)

multiplying (1) with (2) gives

ab = \sqrt{7^2-(5\sqrt{2} )^2} =\sqrt{-1} = -1
(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+3x-14 = 0
(7)

By observation, x = 2 is one of the solution for equation (7)

Then, divide the polynomial x^3+3x-14 by x-2 to find another factor

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

Since x^2+2x+7= (x+1)^2 + 6>0 , there is no real solution for the equation x^2+2x+7 = 0

Therefore, there's only one solution x = 2. Now we conclude

\sqrt{7+5\sqrt{2} } +\sqrt{7-5\sqrt{2} } =2

Steven Zheng posted 5 months ago

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

7+5\sqrt{2}=(1+\sqrt{2})^3

7-5\sqrt{2}=(1-\sqrt{2})^3

Find cube root: \sqrt{7+5\sqrt{2}}=1+\sqrt{2}

Find cube root: \sqrt{7-5\sqrt{2}}=1-\sqrt{2}

Therefore

\sqrt{7+5\sqrt{2}}+\sqrt{7-5\sqrt{2}}=2

Steven Zheng posted 5 months ago

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