| Directions: In each of the following questions two equations are given. Solve these equations and give answer. [IBPS (PO) 2013] |
| I. \[{{x}^{2}}+4x+4=0\] |
| II. \[{{y}^{2}}-8y+16=0\] |
A) If \[x\ge y\]
B) If \[x>y\]
C) If \[x\le y\]
D) If \[x<y\]
E) If \[x=y\] or no relation can be established between x and y
Correct Answer: D
Solution :
| I. \[{{x}^{2}}+4x+4=0\] |
| \[\Rightarrow \]\[{{x}^{2}}+2x+2x+4=0\] |
| \[\Rightarrow \]\[x\,\,(x+2)+2\,\,(x+2)=0\] |
| \[\Rightarrow \] \[(x+2)(x+2)=0\] |
| \[\Rightarrow \] \[{{(x+2)}^{2}}=0\] |
| \[\Rightarrow \] \[x=-\,\,2\] |
| II. \[{{y}^{2}}-8y+16=0\] |
| \[\Rightarrow \]\[{{y}^{2}}-4y-4y+16=0\] |
| \[\Rightarrow \]\[y\,\,(y-4)-4\,\,(y-4)=0\] |
| \[\Rightarrow \] \[(y-4)(y-4)=0\] |
| \[\Rightarrow \] \[{{(y-4)}^{2}}=0\] |
| \[\therefore \] \[y=4\] |
| Hence, \[x<y\] |
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