Question

12
$\frac{12}{?4 \sqrt{3} - \sqrt{2 \} }$
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Answer 1

Answer:

the actual question is

$\frac{12}{4 \sqrt{3} - \sqrt{2} }$

the answer

multiply the fraction by

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

now,

$\frac{12}{4 \sqrt{3} - \sqrt{2} } \times \frac{4 \sqrt{3} + \sqrt{2} }{4 \sqrt{3} + \sqrt{2} }$

to multiply the fractions

multiply numerator and denominator seperately

then,

$\frac{12(4 \sqrt{3} + \sqrt{2} )}{(4 \sqrt{3} - \sqrt{2})(4 \sqrt{3} + \sqrt{2}) }$

using (a - b)(a+b)=a^2-b^2

simplify the product

$\frac{12(4 \sqrt{3} + \sqrt{2} )}{16 \times 3 - 2}$

$\frac{12(4 \sqrt{3} + \sqrt{2} )}{48 - 2}$

$= \frac{12(4 \sqrt{3} + \sqrt{2}) }{46}$

$= \frac{6(4 \sqrt{3} + \sqrt{2} ) }{23}$

$= \frac{24 \sqrt{3} + 6 \sqrt{2} }{23}$

=2.17628

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