Question

$\sqrt{x} + \sqrt{x - \sqrt{1 - x} } = 1$
please give complete steps​

Answer 1

Answer :-

$\implies\sf \sqrt x + \sqrt{x - \sqrt{1-x}} = 1$

Transposing √x on RHS :-

$\implies\sf \sqrt{x - \sqrt{1-x}} = 1 - \sqrt x$

Squaring on both side :-

$\implies\sf (\sqrt{x - \sqrt{1-x}})^2 = (1 - \sqrt x)^2$

$\implies\sf x - \sqrt{1-x} = 1^2 + (\sqrt{x})^2 - 2\sqrt x$

$\implies\sf x - \sqrt{1-x} = 1 + x - 2\sqrt{x}$

$\implies\sf \cancel x - \sqrt{1-x} - \cancel x = 1 - 2\sqrt{x}$

$\implies\sf - \sqrt{1-x} = 1 - 2\sqrt{x}$

Squaring on both side :-

$\implies\sf (- \sqrt{1-x})^2 = (1 - 2\sqrt{x})^2$

$\implies\sf 1 - x = 1^2 + ( 2 \sqrt x )^2 - 4 \sqrt x$

$\implies\sf \cancel 1 - x = \cancel 1 + 4 x - 4 \sqrt x$

$\implies\sf - x = 4x - 4 \sqrt x$

$\implies\sf -x - 4x = - 4\sqrt x$

$\implies\sf -5x = -4\sqrt x$

$\implies\sf 5x = 4 \sqrt x$

Squaring on both side :-

$\implies\sf (5x)^2 = (4 \sqrt x)^2$

$\implies\sf 25x^2 = 16x$

$\implies\sf 25x = 16$

$\implies\boxed{\sf x = \dfrac{16}{25}}$

Answer 2

Step-by-step explanation:

√x-√1-x =1-√x

(√x-√1-x)²= (1-√x)²

x-√-x=(1-√x)²

x-√1-x=(1)²+(√x)²-2*1*√x

x-√1-x=1+x-2√x

x-x-√1-x=1-2√x

Now x will be cancled by x

-√1-x=1-2√x

(-√1-x)²=(1-2√x)²

repeat the same procedure to the right hand side expression and then simplify it further.

1-x=(1)²+2√x²-2* 1*2√x

1-x=1+4x-4√x

1-1-x =4x-4√x

1-1-x=4x-4√x

Now 1 will be cancled by 1

-x=4x-4√x

4√x=4x+x

4√x=5x

4√x²=5x²

16x=25x²

16x-25x²=0

x(25x-16) =0

Therefore,x = 0 or25x-16 = 0  is the solution of this equation.

x = 0 or x=16/25

putting x=0

√x+√x-√1-x=√0+√0-√1-0

0+√0-1

√-1

=>√-1≠1

now,putting x=16/25