Question : If $x+\left [\frac{1}{(x+7)}\right]=0$, what is the value of $x-\left [\frac{1}{(x+7)}\right]$?
Option 1: $3\sqrt{5}$
Option 2: $3\sqrt{5}-7$
Option 3: $3\sqrt{5}+7$
Option 4: $8$
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Correct Answer: $3\sqrt{5}-7$
Solution :
Given: $x+[\frac{1}{(x+7)}]=0$
Adding both sides 7, we get,
⇒ $x+7+[\frac{1}{(x+7)}]=7$
Squaring both sides, we get,
⇒ $(x+7)^{2}+(\frac{1}{x+7})^{2}+2=49$
⇒ $(x+7)^{2}+(\frac{1}{x+7})^{2}=47$
Subtracting 2 from both sides,
$(x+7)^{2}+(\frac{1}{x+7})^{2}–2=47-2$
⇒ $[(x+7)-(\frac{1}{x+7})]^{2}=45$
⇒ $[(x+7)-(\frac{1}{x+7})]=\sqrt{45}$
⇒ $[(x-\frac{1}{x+7})]=\sqrt{45}-7=\sqrt{9 \times5}-7=3\sqrt{5}–7$
Hence, the correct answer is $3\sqrt{5}-7$.
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