A) \[\pm 1\]
B) \[\pm 2\]
C) \[\pm 3\]
D) \[\pm 4\]
Correct Answer: C
Solution :
| [c] Roots of Given equation |
| \[{{x}^{3}}-12{{x}^{2}}+39x-28=0\] |
| are in A.P. |
| Let \[\alpha -\beta ,\] \[\alpha ,\] \[\alpha +\beta \] be the roots of the equation. |
| Sum of the roots |
| \[=\alpha -\beta +\alpha +\alpha +\beta =\frac{-(-12)}{1}=12\] |
| \[3\alpha =12\Rightarrow \alpha =4\] and |
| \[(\alpha -\beta )\alpha +\alpha (\alpha +\beta )+(\alpha +\beta )(\alpha -\beta )=39\] |
| \[\Rightarrow \,\,{{\alpha }^{2}}-\alpha \beta +{{\alpha }^{2}}+\alpha \beta +{{\alpha }^{2}}-{{\beta }^{2}}=39\] |
| \[\Rightarrow \,\,3{{\alpha }^{2}}-{{\beta }^{2}}=39\Rightarrow 3{{(4)}^{2}}-{{\beta }^{2}}=39\] |
| \[\Rightarrow \,\,\,48-{{\beta }^{2}}=39\Rightarrow -{{\beta }^{2}}=39-48\Rightarrow -{{\beta }^{2}}=-9\] |
| \[\Rightarrow \,\,\,\beta =\pm 3\] |
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