A) \[-3\sqrt{2}\]
B) \[8\sqrt{2}\]
C) \[12\sqrt{2}\]
D) \[-2\sqrt{2}\]
Correct Answer: C
Solution :
Equation is \[2{{x}^{2}}+ax+32=0\] Let one root be \[\alpha ,\] then other would be\[2\alpha .\] Now, \[\alpha \times 2\alpha =16\] \[\Rightarrow \] \[\alpha =\pm \,2\sqrt{2}\] and \[\alpha +2\alpha =-a/2\] \[\Rightarrow \] \[3\alpha =-a/2\] \[\Rightarrow \] \[6\alpha =-a\] \[\Rightarrow \] \[\pm \,12\sqrt{2}=-a\] or \[a=\pm 12\sqrt{2}\]
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