Question

Can anyone solve thiss ??
$\sqrt{7 + 4 \sqrt{3} } + \sqrt{7 - 4 \sqrt{3 } } = x$
Find the value of x​

Answer 1

Step-by-step explanation:

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Answer 2

$\sqrt{7 + 4 \sqrt{3} } + \sqrt{7 - 4 \sqrt{3} } = x\\ = > \sqrt{2 {}^{2} + 2.2 \sqrt{3} +( \sqrt{3}) {}^{2} } + \sqrt{2 {}^{2} - 2.2 \sqrt{3} + (\sqrt{ 3} ) {}^{2} } = x \\ = > \sqrt{(2 + \sqrt{3} ) {}^{2} } + \sqrt{(2 - \sqrt{3} ) {}^{2} } = x \\ = > 2 + \sqrt{3} + 2 - \sqrt{3} = x \\ = > x = 4$