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4-2.Quadratic Equations and Inequations
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If the roots of the equation $8{x^3} - 14{x^2} + 7x - 1 = 0$ are in $G.P.$, then the roots are
A
$1,\frac{1}{2},\frac{1}{4}$
B
$2, 4, 8$
C
$3, 6, 12$
D
None of these
Solution
(a) 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.
Standard 11
Mathematics
