Question

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

Accepted Answer

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

Answer

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.

algebra1

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

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