| If \[{{a}^{2}}+{{b}^{2}}+2b+4a+5=0,\]then the value of \[\frac{a-b}{a+b}\]is [SSC (10+2) 2011] |
A) \[3\]
B) \[-\,\,3\]
C) \[\frac{1}{3}\]
D) \[-\frac{1}{3}\]
Correct Answer: C
Solution :
| Given, \[{{a}^{2}}+{{b}^{2}}+2b+4a+5=0\] |
| \[\Rightarrow \]\[{{a}^{2}}+4a+{{b}^{2}}+2b+5=0\] |
| \[\Rightarrow \]\[{{a}^{2}}+4a+4+{{b}^{2}}+2b+1=0\] |
| \[\Rightarrow \]\[{{(a+2)}^{2}}+{{(b+1)}^{2}}=0\] |
| \[[\because {{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab]\] |
| It is possible only, when |
| \[a+2=0\]\[\Rightarrow \]\[a=-\,\,2\] |
| and \[b+1=0\]\[\Rightarrow \]\[b=-\,\,1\] |
| \[\therefore \] \[\frac{a-b}{a+b}=\frac{-\,\,2+1}{-\,\,2-1}=\frac{-\,\,1}{-\,\,3}=\frac{1}{3}\] |
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