Question : If $\frac{22 \sqrt{2}}{4 \sqrt{2}-\sqrt{3+\sqrt{5}}}=a+\sqrt{5} b$, with $a, b>0$, then what is the value of $(a b):(a+b)$?
Option 1: 7 : 8
Option 2: 7 : 4
Option 3: 4 : 7
Option 4: 8 : 7
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Correct Answer: 7 : 8
Solution :
Given: $\frac{22 \sqrt{2}}{4 \sqrt{2}-\sqrt{3+\sqrt{5}}}=a+\sqrt{5} b$
Multiplying and dividing the number $\sqrt{3+\sqrt5}$ by $\sqrt2$
⇒ $\frac{22 \sqrt{2}}{4 \sqrt{2}-(\frac{\sqrt2(\sqrt{3+\sqrt{5})}}{2})}=a+\sqrt{5} b$
⇒ $\frac{22 \sqrt{2}}{4 \sqrt{2}-(\frac{\sqrt{6+2\sqrt{5}}}{2})}=a+\sqrt{5} b$
$\because (\sqrt{5}+1)^2=6+2\sqrt5$
⇒ $\frac{22 \sqrt{2}}{4 \sqrt{2}-(\frac{\sqrt{(\sqrt{5}+1)^2}}{2})}=a+\sqrt{5} b$
⇒ $\frac{22 \sqrt{2}}{4 \sqrt{2}-(\frac{\sqrt{5}+1)}{\sqrt2})}=a+\sqrt{5} b$
⇒ $\frac{22 \sqrt{2}}{(\frac{4 \sqrt{2}\times\sqrt 2-\sqrt{5}-1)}{\sqrt2})}=a+\sqrt{5} b$
⇒ $\frac{22 \sqrt{2}\times\sqrt2}{8-\sqrt{5}-1}=a+\sqrt{5} b$
⇒ $\frac{44}{7-\sqrt{5}}=a+\sqrt{5} b$
On rationalising, we get:
⇒ $\frac{44}{7-\sqrt{5}}\times\frac{7+\sqrt{5}}{7+\sqrt{5}}=a+\sqrt{5} b$
⇒ $\frac{44(7+\sqrt{5})}{44}=a+\sqrt{5} b$
⇒ $7+\sqrt{5}=a+\sqrt{5} b$
So, $a=7, b=1$
Now, $(a b):(a+b)$
= $(7\times1):(7+1)$
= $7:8$
Hence, the correct answer is 7 : 8.
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