First slide

Theory of equations

Question

If the roots of the equation $8 x^{3} - 14 x^{2} + 7 x - 1 = 0$ are in GP, then the roots are

Easy

A

$1, \frac{1}{2}, \frac{1}{4}$
B

2,4,8
C

3, 6 , 12
D

None of these

Solution

Since, the roots of the equation $8 x^{3} - 14 x^{2} + 7 x - 1 = 0$
are in GP. Let the roots $\frac{α}{β}, α, αβ, β \neq 0 .$ Then, the product of roots is $α^{3} = \frac{1}{8}$
$\Rightarrow α = \frac{1}{2} and hence, β = \frac{1}{2}$
So, roots are $1, \frac{1}{2}, \frac{1}{4}$

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