Complex Numbers And Quadratic Equations question 808
Question: If the roots of the equation $ 8x^{3}-14x^{2}+7x-1=0 $ are in G.P., then the roots are [MP PET 1986]
Options:
A) $ 1,\frac{1}{2},\frac{1}{4} $
B) 2, 4, 8
C) 3, 6, 12
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Let the roots be $ \frac{\alpha }{\beta },\alpha ,\alpha \beta ,\beta \ne 0 $ . Then the product of roots is $ {{\alpha }^{3}}=-\frac{-1}{8}=\frac{1}{8}\Rightarrow \alpha =\frac{1}{2} $ and hence $ \beta =\frac{1}{2} $ , so roots are $ 1,\frac{1}{2},\frac{1}{4} $ Trick: By inspection, we get the numbers $ 1,\frac{1}{2},\frac{1}{4} $ satisfying the given equation.
BETA
