A) \[\pm 1\]
B) \[\pm 2\]
C) \[\pm 3\]
D) \[\pm 4\]
Correct Answer: C
Solution :
Let a - d, a, a + d be the roots of the equation \[{{x}^{3}}-12{{x}^{2}}+39x-28=0\] Then \[(a-d)+a+(a+d)=12\] and \[(a-d)\,a\,(a+d)=28\] Þ \[3a=12\]and \[a\,({{a}^{2}}-{{d}^{2}})=28\] Þ \[a=4\] and \[a\,({{a}^{2}}-{{d}^{2}})=28\] Þ \[16-{{d}^{2}}=7\Rightarrow d=\pm \,3\].
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