# explain the methodmy tutor said that is the raise to square method! what is this? square root(1-square root(x^4-x))+1=x

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We have to find the solution for sqrt(1-sqrt(x^4-x))+1=x

sqrt(1-sqrt(x^4-x))+1=x

sqrt(1-sqrt(x^4-x)) = x - 1

take the square of both the sides

- sqrt(x^4 - x) = x^2 - 2x

take the square of both the sides again

=> x^4 - x = x^4 + 4x^2 - 4x^3

=> - x = 4x^2 - 4x^3

=> 4x^3 - 4x^2 - x = 0

=> x(4x^2 - 4x - 1) = 0

x1 = 0

x2 = [4 + sqrt(16 + 16)]/8

=> 1/2 + 4*sqrt 2/8

=> 1/2 + sqrt 2/2

x3 = 1/2 - sqrt 2/2

**The solutions are { 0, 1/2 + 1/sqrt 2, 1/2 - 1/sqrt 2}**

### We'll move the number alone to the right side to keep only the square root to the left:** **

**sqrt[1-sqrt(x^4-x)]= (x-1)**

**Now, raising to square both sides, we'll eliminate the square root from the left side:**

1-square root(x^4-x)= (x-1)^2

We'll eliminate like terms:

1-square root(x^4-x)= x^2 - 2x + 1 -square root(x^4-x)= x^2 - 2x

We'll raise to square both sides again: x^4 - x = x^4 - 4x^3 + 4x^2

We'll eliminate x^4 both sides: 4x^3 - 4x^2 - x = 0

We'll factorize by x: x(4x^2 - 4x - 1) = 0

We'll put x = 0. 4x^2 - 4x - 1 = 0 We'll apply quadratic formula:

x1 = [4+sqrt(16+16)]/16 x1 = 4(1 + sqrt2)/16 x1 = (1 + sqrt2)/4 x2 = (1 - sqrt2)/4

Since the range of admissible values for x is: {0}U[(1 + sqrt2)/4 ; +infinite), we'll reject the second value for x.

**The solutions of the equation are: {0 ; (1 + sqrt2)/4 }.**