# square root(x+square root(1-x))+square rootx=1. What is x?

*print*Print*list*Cite

### 2 Answers

We have to solve the equation sqrt (x + sqrt (1 - x)) + sqrt x = 1

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

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

square both the sides

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

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

square both the sides

=> 1 - x = 1 + 4x - 4 sqrt x

=> 5x = 4 sqrt x

square both the sides

25x^2 = 16x

=> 25x^2 - 16x = 0

=> x(25x - 16) = 0

=> x = 0 and x = 16/25

**The required values of x are 0 and 16/25**

First, we'll impose the constraints of existence of square roots:

x>=0

1 - x>=0

x=<1

The range of admissible values for x is [0 ; 1].

We'll solve the equation, moving sqrt x to the right side:

sqrt[x+sqrt(1-x)] = 1 - sqrtx

We'll raise to square both sides:

x + sqrt(1-x) = 1 - 2sqrtx + x

We'll eliminate x both sides:

sqrt(1-x) = 1 - 2sqrtx

We'll raise to square again:

1 - x = 1 - 4sqrtx + 4x

We'll eliminate 1:

4sqrtx - 4x - x = 0

4sqrtx - 5x = 0

4sqrtx = 5x

We'll raise to square both sides, to eliminate the square root:

16x = 25x^2

We'll subtract 16 x:

25x^2 - 16x = 0

We'll factorize by x:

x(25x - 16) = 0

x1 = 0

25x = 16

x2 = 16/25

Since both values of x are in the range of admissible values, we'll accept them as solutions of the given equation.

**x1 = 0 and x2 = 16/25.**