Question

Another Question,
Solve the given equation.
$\sf \sqrt{5 - x} = 5 - {x}^{2}$
Please don't give wrong or irrelevant answer. It's my request.
Best answer will be marked as brainliest. ​

Answer 1

Hey mate,

Given :-

$\sf \sqrt{5 - x} = 5 - {x}^{2}$

Solving :-

Find the value of x

Answer :-

$\sf \sqrt{5} {}^{}$

Explanation :-

($\sf \sqrt{5 + x} {}^{}$) - $\sf \sqrt{5x} {}^{2}$) = 0

=> ($\sf \sqrt{5 - x} {}^{}$) - ($\sf \sqrt{5 + x} {}^{}$) ($\sf \sqrt{5 - x} {}^{}$) = 0

=> ($\sf \sqrt{5 - x} {}^{}$) ( 1 - $\sf \sqrt{5} {}^{}$ + x) = 0

Hence , the value of x is $\sf \sqrt{5} {}^{}$

Hope it helps...

Answer 2

Answer:

Hey mate,

Given :-

\sf \sqrt{5 - x} = 5 - {x}^{2}

5−x

=5−x

2

Solving :-

Find the value of x

Answer :-

\sf \sqrt{5} {}^{}

5

Explanation :-

(\sf \sqrt{5 + x} {}^{}

5+x

) - \sf \sqrt{5x} {}^{2}

5x

2

) = 0

=> (\sf \sqrt{5 - x} {}^{}

5−x

) - (\sf \sqrt{5 + x} {}^{}

5+x

) (\sf \sqrt{5 - x} {}^{}

5−x

) = 0

=> (\sf \sqrt{5 - x} {}^{}

5−x

) ( 1 - \sf \sqrt{5} {}^{}

5

+ x) = 0

Hence , the value of x is \sf \sqrt{5} {}^{}

5

Hope it helps...

Please brainlest me the answer