A) \[a=-7,\,\,b=-1\]
B) \[a=5,\,\,b=-1\]
C) \[a=2,\,b=-6\]
D) \[a=0,\,b=-6\]
Correct Answer: D
Solution :
| [d] Let \[p(x)={{x}^{2}}+(a+1)x+b\] |
| Given that 2 and \[-3\]are the zeroes of the quadratic polynomial \[p(x)\]. |
| \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,p(2)=0\] and \[p(-3)=0\] |
| Now, \[p(2)=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{2}^{2}}+(a+1)\,(2)+b=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2a+b=-6\] .(1) |
| Also, \[p(-3)=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,{{(-3)}^{2}}+(a+1)\,(-3)+b=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,3a-b=6\] ..(2) |
| On solving eqs. (1) and (2), we get |
| \[a=0,\,\,b=-6\] |
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