Quantitative Aptitude Ques 1902
Question: If $ x=3+2\sqrt{2}, $ then $ \frac{x^{6}+x^{4}+x^{2}+1}{x^{3}} $ is equal to
Options:
A) 216
B) 192
C) 198
D) 204
Show Answer
Answer:
Correct Answer: D
Solution:
- Given, $ x=3+2\sqrt{2}=3+\sqrt{8} $
and $ \frac{1}{x}=\frac{1}{3+\sqrt{8}}\times \frac{3-\sqrt{8}}{3-\sqrt{8}} $
$ =\frac{3-\sqrt{8}}{9-8}=(3-\sqrt{8}) $
Now, $ \frac{x^{6}+x^{4}+x^{2}+1}{x^{3}}=\frac{x^{3}( x^{3}+x+\frac{1}{x}+\frac{1}{x^{3}} )}{x^{3}} $
$ =x(x^{2}+1)+\frac{1}{x}( \frac{1}{x^{2}}+1 ) $
(i)
Now, putting the values of x and $ \frac{1}{x} $ in Eq. (i), we get
$ =(3+\sqrt{8})\times [{{(3+\sqrt{8})}^{2}}+1] $
$ +(3-\sqrt{8})[{{(3-\sqrt{8})}^{2}}+1] $
$ =(3+\sqrt{8})(9+8+6\sqrt{8}+1) $
$ +(3-\sqrt{8})(9+8-6\sqrt{8}+1) $
$ =(3+\sqrt{8})(18+6\sqrt{8})+(3-\sqrt{8})(18-6\sqrt{8}) $
$ =(3+\sqrt{8})6,(3+\sqrt{8})+(3-\sqrt{8})6(3-\sqrt{8}) $
$ =6{{(3+\sqrt{8})}^{2}}+6{{(3-\sqrt{8})}^{2}} $
$ =6[{{(3+\sqrt{8})}^{2}}+{{(3-\sqrt{8})}^{2}}] $
$ =6\times [9+8+6\sqrt{8}+9+8-6\sqrt{8}] $
$ =6\times 34=204 $
BETA
