A) \[{{x}^{2}}-x+12\]
B) \[{{x}^{2}}+x+12\]
C) \[\frac{{{x}^{2}}}{2}-\frac{x}{2}-6\]
D) \[2{{x}^{2}}+2x-24\]
Correct Answer: C
Solution :
| [c] Let the zeroes of a quadratic polynomial are \[\alpha =-3\] and \[\beta =4.\] |
| Then, sum of zeroes \[=\alpha +\beta =-3+4=1\] |
| Product of zeroes \[=\alpha \beta =-3\times 4=-12\] |
| \[\therefore \] Required polynomial |
| \[={{x}^{2}}-(sum\,\,of\,\,zeroes)x+(product\,\,of\,zeroes)\]\[={{x}^{2}}-(1)x+(-12)={{x}^{2}}-x-12\] i.e., \[\frac{{{x}^{2}}}{2}-\frac{x}{2}-6\] |
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