Question

find the value of determinant 3 root 6 minus 4 root 2 upon 5 root 3 minus x ​

Answer 1

$determinant = d = \sqrt{ {b}^{2} - 4ac } \\ 3 \sqrt{6} {x}^{2} - x - \frac{4 \sqrt{2} }{5 \sqrt{3} } \\ d = \sqrt{1 -4 \times 3 \sqrt{6 } \times ( \frac{ - 4 \sqrt{2} }{5 \sqrt{3} } } ) \\ d = \sqrt{1 + \frac{48 \sqrt{12} }{5 \sqrt{3} } } \\ d= \sqrt{1 + \frac{48 \sqrt{4} \times \sqrt{3} }{5 \sqrt{3} } } \\ d = \sqrt{1 + \frac{96}{5} } = \sqrt{ \frac{5 + 96}{5} } = \sqrt{ \frac{101}{5} }$

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Answer 2

Step-by-step explanation:

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