Question

If
$\sqrt{19 - 4 \sqrt{x} } = \sqrt{12} - \sqrt{7}$
then value of x is?​

Answer 1

AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

• An Equation in one variable
• $\tt \sqrt{19 - 4 \sqrt{x} } = \sqrt{12} - \sqrt{7}$

${\sf{\green{\underline{\large{To\:find}}}}}$

• Value of x

${\sf{\pink{\underline{\Large{Explanation}}}}}$

$\tt \sqrt{19 - 4 \sqrt{x} } = \sqrt{12} - \sqrt{7}$

Both side squaring

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$\tt (\sqrt{19 - 4 \sqrt{x} } )^2=( \sqrt{12} - \sqrt{7} )^2$

$\tt\implies(\sqrt{19 - 4 \sqrt{x} } )^2=( \sqrt{12} - \sqrt{7} )^2$

$\tt\implies{19 - 4 \sqrt{x}} =12+7+2\times \sqrt{12}\times\sqrt{7}$

$\tt\implies{19-4\sqrt{x}} =19+2\times\sqrt{12}\times\sqrt{7}$

$\tt\implies{\cancel{19}-4\sqrt{x}} =\cancel{19}+2\times\sqrt{12}\times\sqrt{7}$

$\tt\implies{-4\sqrt{x}} =+2\times\sqrt{84}$

$\tt\implies({-4\sqrt{x}})^2=+(2\times\sqrt{84})^2$

$\tt\implies{16x}=4\times84$

$\tt\implies{x}=\cancel{\frac{4\times84}{\cancel{16}}}$

$\tt\therefore{X}=21$

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Answer 2

Answer:

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