How to solve the equation x^8 - 64 = 0?

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to solve x^8 - 64 = 0

Now 64 = 2^6

x^8 - 64 = 0

=> x^8 = 64

=> x^8 = 2^6

=> x = 2^(6/8)

=> x = 2^(3/4)

=> x = 8^(1/4)

=> x=  sqrt [2*sqrt 2] and x = -sqrt [2*sqrt 2]

The values of x are x=sqrt [2*sqrt 2] and x=-sqrt [2*sqrt 2]

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

Let's recall first the formula for the difference of squares:

a^2 - b^2 =(a-b)(a+b)

Now,we'll put a^2 = x^8 = (x^4)^2 and b^2 = 64 = 8^2

We'll re-write the given equation, emphasizing on the difference of squares:

 (x^4)^2 - 8^2 = (x^4 - 8)(x^4 + 8)

x^4 - 8 is also a difference o squares, whose terms are: a = x^2 and b = 2sqrt2

(x^4 - 8)(x^4 + 8) = (x^2 - 2sqrt2)(x^2 + 2sqrt2)(x^4 + 8)

Now, we'll solve the equation:

x^8 - 64 = 0

x^8 - 64 = (x^2 - 2sqrt2)(x^2 + 2sqrt2)(x^4 + 8)

(x^2 - 2sqrt2)(x^2 + 2sqrt2)(x^4 + 8) = 0

x^2 - 2sqrt2 = 0

x^2 = 2sqrt2

x1 = +sqrt(2sqrt2)

x2 = -sqrt(2sqrt2)

x^2 + 2sqrt2 = 0

x3 = +i*sqrt(2sqrt2)

x4 = -i*sqrt(2sqrt2)

The only real roots of the equation are {+sqrt(2sqrt2) ; -sqrt(2sqrt2)}.

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