Question

√3x2 -2√2x-2√3 anyone solved by quadratic formula​

Answer 1

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

Explanation:

Given,

$\sqrt{3} {x}^{2} - 2 \sqrt{2} x - 2 \sqrt{3}$

Here,

$a = \sqrt{3} \\ \: \: \: \: b = - 2 \sqrt{2} \\ \: \: \: \: c = - 2 \sqrt{3}$

Therefore,

$x = \frac{ - b \frac{ + }{} \sqrt{ {b}^{2} - 4ac} }{2a}$

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

$\: \: \: = \frac{2 \sqrt{2} \frac{ + }{} \sqrt{(4 \times 2) + (8 \times 3)} }{2 \sqrt{3} }$

$\: \: \: = \frac{2 \sqrt{2} \frac{ + }{} \sqrt{8 + 24} }{2 \sqrt{3} }$

$\: \: \: = \frac{2 \sqrt{2} \frac{ + }{} \sqrt{32} }{2 \sqrt{3} }$

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

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

$\: \: \: = \frac{ \sqrt{2} \frac{ + }{}2 \sqrt{2} }{ \sqrt{3} }$

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