A) 12
B) \[10\]
C) \[7\]
D) \[1\]
Correct Answer: B
Solution :
[b] It is given that 4 and \[-3\] are the zeroes of the polynomial |
\[{{x}^{2}}+(a+1)x+b.\] |
\[\therefore \] Sum of zeroes \[=-\frac{(a+1)}{1}=-a-1\] |
\[\Rightarrow \,\,4+(-3)=-a-1\Rightarrow 1=-a-1\Rightarrow a=-2\] |
And product of zeroes = b |
\[\Rightarrow \,\,\,\,\,\,\,\,(4)\,(-3)=b\Rightarrow \,=-12\] |
\[\therefore \,\,\,\,\,\,\,a-b=-2\,(-12)=-2+12=10\] |
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