# Solve the equation for x square root(1-square root(x^4-x))=x-1

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We have to solve: sqrt(1 - sqrt (x^4 - x)) = x - 1

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

square both the sides

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

=> (1 - sqrt (x^4 - x)) = (x^2 + 1 - 2x)

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

square both the sides again

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

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

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

x1 = 0

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

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

=> 1/2 + 1/sqrt 2

The other root of the quadratic equation is not a valid solution.

**The values are x = 0 and x = 1/2 + 1/sqrt 2**

First, we'll raise to square both sides, to 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 }.**