Question

If the roots of the equation $x^3 - 7x^2 + 14x - 8 = 0$ are in geometric progression, then the common ratio can be:

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

If the roots of a cubic equation are in GP, the middle term $a$ is always the cube root of the constant term (with sign changed if the leading coefficient is 1). Here, $\sqrt[3]{8} = 2$.

Answer 1

The Correct Option is B

Step 1: Concept
Let the roots of a cubic equation $ax^3 + bx^2 + cx + d = 0$ be in geometric progression (GP). We can represent these roots as $\frac{a}{r}$, $a$, and $ar$, where $r$ is the common ratio.

Step 2: Meaning
The product of the roots of the cubic equation is given by $-d/a$. Here, the equation is $x^3 - 7x^2 + 14x - 8 = 0$, so the product of the roots is $8$.

Step 3: Analysis
Product of roots: \[ \left(\frac{a}{r}\right) \cdot a \cdot (ar) = 8 \implies a^3 = 8 \implies a = 2 \] Sum of roots: \[ \frac{a}{r} + a + ar = 7 \implies 2\left(\frac{1}{r} + 1 + r\right) = 7 \implies \frac{1}{r} + 1 + r = \frac{7}{2} \] \[ \implies r + \frac{1}{r} = \frac{5}{2} \implies 2r^2 - 5r + 2 = 0 \implies (2r - 1)(r - 2) = 0 \implies r = 2 \text{ or } r = \frac{1}{2} \]

Step 4: Conclusion
Thus, the possible common ratio is $2$. Final Answer: (B)