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: Smart way to name GP roots.
When three numbers are in GP, call them $\frac{a}{r}$, $a$, $ar$. This makes their product very clean.

Step 2: Use the product of roots.
For $x^3 - 7x^2 + 14x - 8 = 0$, the product of roots is $8$. So \[ \frac{a}{r} \cdot a \cdot ar = a^3 = 8 \implies a = 2 \] So one root is exactly 2.

Step 3: Use the sum of roots.
The sum of roots is $7$: \[ 2\left(\frac{1}{r} + 1 + r\right) = 7 \]

Step 4: Simplify.
Divide by 2: \[ r + \frac{1}{r} = \frac{5}{2} \]

Step 5: Clear the fraction.
Multiply by $2r$: \[ 2r^2 - 5r + 2 = 0 \]

Step 6: Solve.
Factor: $(2r-1)(r-2)=0$, so $r = 2$ or $r = \tfrac12$. The whole-number ratio in the options is \[ \boxed{ r = 2 } \]