Question

the value of
$\sqrt{32} + \sqrt{48} \div \sqrt{8} - \sqrt{12}$
​

Answer 1

Answer:

$\sqrt{32} + \sqrt{48} \div \sqrt{8} - \sqrt{12}$

$= 4 \sqrt{2} + 4 \sqrt{3} \div 2 \sqrt{2} - 2 \sqrt{3}$

$= \frac{4 \sqrt{2} + 4 \sqrt{3} }{2 \sqrt{2} - 2 \sqrt{3} }$

$= \frac{4( \sqrt{2} + \sqrt{3} )}{2( \sqrt{2} - \sqrt{3} }$

$= \frac{2( \sqrt{2} + \sqrt{3} )}{ \sqrt{2} - \sqrt{3} } \times \frac{ \sqrt{2} + \sqrt{3} }{ \sqrt{2} + \sqrt{3} }$

$= 4 \sqrt{6} - 10$

Answer 2

Answer:

$\sqrt{32} + \sqrt{48} \div \sqrt{8} - \sqrt{12}$

$= 4 \sqrt{2} + 4 \sqrt{3} \div 2 \sqrt{2} - 2 \sqrt{3}$

$= \frac{4 \sqrt{2} + 4 \sqrt{3} }{2 \sqrt{2} - 2 \sqrt{3} }$

$= \frac{4( \sqrt{2} + \sqrt{3} )}{2( \sqrt{2} - \sqrt{3} }$

$= \frac{2( \sqrt{2} + \sqrt{3} )}{ \sqrt{2} - \sqrt{3} } \times \frac{ \sqrt{2} + \sqrt{3} }{ \sqrt{2} + \sqrt{3} }$

= -10 + 4√6