| If \[x=\frac{\sqrt{2}+1}{\sqrt{2}-1}\]and \[x-y=4\sqrt{2},\]then the value of \[({{x}^{2}}+{{y}^{2}})\]is [SSC(CPO) 2013] |
A) 34
B) 38
C) 30
D) 32
Correct Answer: A
Solution :
| \[x=\frac{\sqrt{2}-1}{\sqrt{2}-1}\]\[\Rightarrow \]\[x=\frac{\sqrt{2}+1}{\sqrt{2}-1}\times \frac{\sqrt{2}+1}{\sqrt{2}+1}\] |
| \[\Rightarrow \] \[x=\frac{2+1+2\sqrt{2}}{1}\] |
| \[\Rightarrow \] \[x=3+2\sqrt{2}\] (i) |
| and \[x-y=4\sqrt{2}\] |
| \[\Rightarrow \] \[y=x-4\sqrt{2}\] |
| \[=3+2\sqrt{2}-4\sqrt{2}\] [from Eq. (i)] |
| \[=3-2\sqrt{2}\] |
| Now, \[{{x}^{2}}+y={{(3+2\sqrt{2})}^{2}}+{{(3-2\sqrt{2})}^{2}}\] |
| \[=9+8+12\sqrt{2}+9+8-12\sqrt{2}=34\] |
You need to login to perform this action.
You will be redirected in
3 sec
