Question

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

Answer 1

Answer:

$→ \:x = \dfrac{16}{25}$

Step-by-step explanation:

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

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Answer 2

Answer:

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

Step-by-step explanation:

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