Question

if$\sqrt{x} +\sqrt{x-\sqrt{1-x}}=1$ then the value of x is​

Answer 1

Answer:

$→ \:x = \dfrac{16}{25}→x=2516$

Step-by-step explanation:

$\begin{gathered}→ \:\sqrt{x} \: + \: \sqrt{x - \sqrt{1 - x} } \: = \: 1 \\ \\ → \:\sqrt{x - \sqrt{1 - x} } \: = \: 1 - \sqrt{x } \\ \\ {\Large{Squaring \: On \: Both \: Sides}} \\ \\ →\: \bigg( \: \: \sqrt{x - \sqrt{1 - x} } \: \: \: \: \bigg){\large{^{^{2}}}} \: = \: ( \: 1 - \sqrt{x} \: ) {}^{2} \\ \\ →\: x \: - \: \sqrt{1 - x} \: = \: 1 \: + \: x \: - \: 2 \sqrt{x} \\ \\→\: - \sqrt{1 - x} \: = \: 1 \: - \: 2 \sqrt{x} \\ \\→\: ( \: - \sqrt{1 - x} \: ) {}^{2} \: = \: ( \: 1 - 2 \sqrt{x} \: ) {}^{2} \\ \\→\: 1 \: - \: x \: = \: 1 \: + \: 4x \: - \: 4 \sqrt{x} \\ \\→\: 4 \sqrt{x } \: = \: 5x \\ \\→\: (4 \sqrt{x} ) {}^{2} \: = \: (5x) {}^{2} \\ \\→\: 16 \: . \: x \: = \: 25 \: x {}^{2} \\ \\→\: x \: = \: \dfrac{16}{25}\end{gathered}$

hope help u mate ✌

Answer 2

Step-by-step explanation:

√x + √(x - √(1 - x)) = 1

√(x - √(1 - x)) = 1 - √x

Squaring both sides,

→ [√(x - √(1 - x))]² = (1 - √x)²

→ x - √(1 - x) = 1 + x - 2√x

→ √(1 - x) = 1 + x - 2√x - x

→ √(1 - x) = 1 - 2√x

Again squaring both sides,

→ [√(1 - x)]² = (1 - 2√x)²

→ 1 - x = 1 + 4x - 4√x

→ 0 = 1 - 1 + 4x +x - 4√x

→ 0 = 5x - 4√x

→ 4√x = 5x

Do squaring both sides,

→ (4√x)² = (5x)²

→ 16x = 25x²

x throughout cancel,

→ 16 = 25x

→ x = 16/25

Hence, the value of x is 16/25.