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Sveta_85 [38]
2 years ago
8

Solve for x -

Mathematics
2 answers:
Vanyuwa [196]2 years ago
7 0
Too much to write ... so I have hand written the answer. 

weqwewe [10]2 years ago
5 0
QUADRATIC \: \: RESOLUTIONS \\ \\ \\ Given \: \:expression - \\ \\ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{a + b + x} \\ \\ \frac{1}{a} + \frac{1}{b} = \frac{1}{a + b + x} - \frac{1}{x} \\ \\ \frac{a + b}{ab} = \frac{ - (a + b)}{x(a + b + x)} \\ \\ \frac{1}{ab} = \frac{ - 1}{x(a + b + x)} \\ \\ Simplifying \: \: above \: \: expression \: \: , \: \\ \\ We \: get \: a \: Quadratic \: equation \: - \\ \\ {x}^{2} + (a + b)x + ab = 0 \\ \\ (x + a)(x + b) = 0 \\ \\ \\ || \: \: \: x = - a \: \: \: || \: \: \: x = - b \: \: \: || \: \: \: \: \: \: \: \: \: \: \: Ans.
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Work out the following and give your answer in it's simplest form:
Pavel [41]

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3 years ago
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<h3>Used:</h3>

\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\ \text{for}\ a\geq0\ \wedge\ b\geq0\\\\\sqrt{a}=b\iff b^2=a\ \text{for}\ a\geq0\ \wedge\ b\geq0\\\\\sqrt{16}=4\ \text{because}\ 4^2=16

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