A) \[-2,-5\]
B) \[-3,-4\]
C) \[2,5\]
D) \[3,4\]
Correct Answer: B
Solution :
[b] Let \[f(x)={{x}^{2}}+7x+12\] |
\[={{x}^{2}}+4x+3x+12\] |
\[=x(x+4)+3(x+4)=(x+4)\,(x+3)\] |
For zeroes of \[f(x),\] |
\[(x+4)\,\,(x+4)=0\] |
\[x+4=0\] or \[x+3=0\] |
\[\Rightarrow \,\,\,\,\,\,\,\,\,\,x=-4,-3\] |
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