Question

please anyone solve the value of x....i need the full answer ​

Answer 1

Given:-

• $\sqrt{x} + \sqrt{x - \sqrt{1 - x} } = 1$

To find:-

• The value of x.

Answer:-

Given that

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

$\longrightarrow \sqrt{x - \sqrt{1 - x} } = 1 - \sqrt{x}$

Squaring both sides,

${ \longrightarrow {\Big(\sqrt{x - \sqrt{1 - x} } \Big)}^{2} = {\Big(1 - \sqrt{x} \Big)}^{2} }$

Expanding R.H.S. using (a - b)² = a² + b² - 2ab with a = 1 and b = √x,

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

${\longrightarrow - \sqrt{1 - x} = 1 - 2\sqrt{x}}$

Squaring both sides,

${\longrightarrow { \big( - \sqrt{1 - x} \big)}^{2} = { \big(1 - 2\sqrt{x}} \big)}^{2}$

Expanding R.H.S. using (a - b)² = a² + b² - 2ab with a = 1 and b = 2√x,

${ \displaystyle\longrightarrow 1 - x = {1}^{2} + { (2 \sqrt{x}) }^{2} - 2(1)(2 \sqrt{x} ) }$

${ \displaystyle\longrightarrow 1 - x = 1 + 4x - 4 \sqrt{x} }$

${ \displaystyle\longrightarrow 4 \sqrt{x} = 4x + x }$

${ \displaystyle\longrightarrow 4 \sqrt{x} = 5x }$

${ \displaystyle\longrightarrow \dfrac{ x }{ \sqrt{x} } = \dfrac{4}{5} }$

${ \displaystyle\longrightarrow \sqrt{x} = \dfrac{4}{5} }$

Squaring both sides,

$\longrightarrow \underline{ \underline{x = \dfrac{16}{25}}}$