If \(x\) is a positive number and \(\sqrt{\frac{3}{2}+{\sqrt{\frac{3}{2}+\sqrt{\frac{3}{2}+\sqrt{\frac{3}{2}+...}}}}}=x\), where the given expression extends to an infinite number of roots, then what is the value of \(x\)?
Two ways..
If you are aware of the type of these questions..\(\sqrt{\frac{3}{2}+{\sqrt{\frac{3}{2}+\sqrt{\frac{3}{2}+\sqrt{\frac{3}{2}+...}}}}}=x\)..
We can further write this as
\(\sqrt{\frac{3}{2}+x}=x\)
Square both sides...\(\frac{3}{2}+{x}=x^2......x^2-x-\frac{3}{2}=0\)
Apply the formula for roots, \(= \frac{-b ±\sqrt{b^2- 4*a*c}}{2a}=\frac{1^2 ± \sqrt{(-1)^2+4*\frac{3}{2}}}{2} = \frac{1 ± \sqrt{1+6}}{2}\)
But x is surely positive, so \(x= \frac{1 + \sqrt{7}}{2} \)
Next, if you do not know anything but are looking for a start...Now 3/2=1.5
what is \(\sqrt{3/2}=\sqrt{1.5}>1\)
\(\sqrt{1.5+{\sqrt{1.5}}<x........x>\sqrt{1.5+1}=\sqrt{2.5}\), so x is surely >1.5
Look at the choices that gives you these values..
A. \(\frac{1-\sqrt{7}}{2}\)... a NEGATIVE value...N)
B. \(\frac{1}{2}\)=0.5...NO
C. \(\frac{\sqrt{7}-1}{2}\)....\(\sqrt{7}\) is between 2 and 3. Even if it is 3, \(\frac{3-1}{2}=1\)..NO
D. \(\frac{\sqrt{7}}{2}\)....Surely \(<\frac{3}{2}...<1.5\)....NO
E. \(\frac{1+\sqrt{7}}{2}\)....Surely \(<\frac{1+3}{2}...<2\)......Possible
So, Only E is possible
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