A) \[-36\]
B) \[22\]
C) \[-\sqrt{35}\]
D) \[\sqrt{21}\]
Correct Answer: A
Solution :
(a): \[{{x}^{3}}+9{{x}^{2}}+6x+c\] Let roots be (a ? d), a and (a + d) Sum of roots \[\alpha +\beta +\gamma =\frac{-b}{a}=-9\] i.e. \[\left( ad \right)+a+\left( a+d \right)=-9\] \[\Rightarrow 3a=-9\] \[a=-3\] Sum product of roots taken two at a time:\[\left( a-d \right)a+\left( a+d \right)a+\left( a-d \right)\left( a+d \right)\] \[={{a}^{2}}-ad+{{a}^{2}}+ad+{{a}^{2}}-{{d}^{2}}\] \[=3{{a}^{2}}-{{d}^{2}}=3{{(-3)}^{2}}-{{d}^{2}}=27-{{d}^{2}}=\frac{c}{a}=6\] \[=27-{{d}^{2}}=\frac{c}{a}=6\] \[\Rightarrow {{d}^{2}}=21\] Product of roots = \[\left( a-d \right)\text{ }a\text{ }\left( a+d \right)\] \[=a\left( {{a}^{2}}-{{d}^{2}} \right)\] \[=a\left( 9-{{d}^{2}} \right)=a(9-21)\] \[=-\,3\,(-12)=36\] Also product of roots \[=\frac{-c}{1}=36\] \[\Rightarrow c=-36\]
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