Question #8a660

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Question #8a660

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Answer 1 · 2017-10-29T12:31:57

$x = 12.2 + \sqrt[2]{7} i$ $x= 12.2 +root2 7 i$
$x = 12.2 - \sqrt[2]{7} i$ $x= 12.2 -root2 7 i$

Explanation:

$- 2 \sqrt[2]{x - 12.2} - 4 = 10$ $-2root2 (x-12.2) -4 = 10$

Hence;
$\sqrt[2]{x - 12.2} + 2 = - 5$ $root2 (x-12.2) +2 = -5$
$\sqrt[2]{x - 12.2} = - 7$ $root 2 (x-12.2) = -7$

Hence;
$x - 12.2 = \sqrt[2]{- 7}$ $x-12.2 = root2 (-7)$
Note $\sqrt[2]{- 7} = \sqrt[2]{- 1} \cdot \sqrt[2]{7}$ $root2 (-7) = root2 (-1) * root2 7$
Hence $x - 12.2 = \sqrt[2]{7} \cdot i$ $x-12.2 = root2 7 *i$

$x = 12.2 + \sqrt[2]{7} i$ $x= 12.2 +root2 7 i$
$x = 12.2 - \sqrt[2]{7} i$ $x= 12.2 -root2 7 i$

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Question #8a660

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