The real values of a for which the quadratic equation 2x2−a3+8a−1x+a2−4a=0 possesses roots of opposite signs are given by
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a
a>5
b
0
c
a>0
d
a>7
answer is B.
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Detailed Solution
The roots of the given equation will be of opposite signs if they are real and their product is negative. i.e. Discriminant ≥ 0 and Product of roots <0⇒ a3+8a−12−8a2−4a≥0 and a2-4a2<0⇒ a2−4a<0∵a2−4a<0⇒a3+8a−12−8a2−4a≥0⇒ 0