A) \[{{x}^{2}}-\frac{1}{6}x+2\]
B) \[{{x}^{2}}+\frac{1}{6}x-2\]
C) \[6{{x}^{2}}-x+12\]
D) \[6{{x}^{2}}+x-12\]
Correct Answer: B
Solution :
[b] It is given that, sum of zeroes \[=-\frac{1}{6}\] and product of zeroes \[=-2\]. Hence, the required polynomial is \[{{x}^{2}}-(\text{sum of zeroes)}\,\text{x+}\,\text{(product of zeroes})\] |
\[={{x}^{2}}-\left( -\frac{1}{6} \right)x+(-2)\]i.e., \[{{x}^{2}}+\frac{1}{6}x-2\] |
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