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 |
\[(\alpha -\beta )\alpha +\alpha (\alpha +\beta )+(\alpha +\beta )(\alpha -\beta )=39\] |
\[\Rightarrow \,\,{{\alpha }^{2}}-\alpha \beta +{{\alpha }^{2}}+\alpha \beta +{{\alpha }^{2}}-{{\beta }^{2}}=39\] |
\[\Rightarrow \,\,3{{\alpha }^{2}}-{{\beta }^{2}}=39\Rightarrow 3{{(4)}^{2}}-{{\beta }^{2}}=39\] |
\[\Rightarrow \,\,\,48-{{\beta }^{2}}=39\Rightarrow -{{\beta }^{2}}=39-48\Rightarrow -{{\beta }^{2}}=-9\] |
\[\Rightarrow \,\,\,\beta =\pm 3\] |
You need to login to perform this action.
You will be redirected in
3 sec