Sequence And Series Question 405
Question: If the roots of the equation $ x^{3}-12x^{2}+39x-28=0 $ are in A.P., then their common difference will be:
Options:
A) $ \pm 1 $
B) $ \pm 2 $
C) $ \pm 3 $
D) $ \pm 4 $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Roots of Given equation $ x^{3}-12x^{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 $
BETA
