A) both positive
B) both negative
C) one positive and other negative
D) both equal
Correct Answer: B
Solution :
Let \[f(x)={{x}^{2}}+20x+96\] |
[\[96=12\times 8=6\times 16=2\times 48.....\] |
\[\therefore \]Here, we will take 12 and 8 as a factors of 96. |
So, middle term is \[20=12+8\]] |
\[={{x}^{2}}+12x+8x+96\] |
\[=x(x+12)+8(x+12)\] |
\[=(x+12)\,\,(x+8)\] |
For the zeroes of \[f(x),\] |
\[x+12=0\] or \[x+8=0\] |
\[\therefore \,\,\,\,x=-8,-12\] |
i.e., both zeroes are negative |
So, option [b] is correct. |
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