A) \[a=-7,\,b=-1\]
B) \[a=5,\,b=-1\]
C) \[a=2,\,b=-6\]
D) \[a=0,\,b=-6\]
Correct Answer: D
Solution :
| Let \[p\left( x \right)={{x}^{2}}+\left( a+1 \right)x+b\] |
| Given that, 2 and -3 are the zeroes of the quadratic polynomial \[p\left( x \right)\] |
| \[\therefore \,\,\,p\left( 2 \right)=0\] and \[p\left( -3 \right)=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,{{2}^{2}}+\left( a+1 \right)\left( 2 \right)+b=0\] |
| \[\Rightarrow \,\,\,\,\,\,4+2a+2+b=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,2a+b=-6\] (i) |
| And \[{{\left( -3 \right)}^{2}}+\left( a+1 \right)\left( -3 \right)+b=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,9-3a-3+b=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,3a-b=6\] (ii) |
| On adding Eqs. (i) and (ii), we get |
| \[5a=0\Rightarrow a=0\] |
| Put the value of a in Eq. (i), we get |
| \[2\times 0+b=-6\] |
| \[\Rightarrow \,\,\,\,\,b=-6\] |
| So, the required values are a = 0 and \[b=-6\] |
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