Question

If $\alpha, \beta, \gamma$ are the roots of the equation $8x^3 - 42x^2 + 63x - 27 = 0$, If $\beta<\gamma<\alpha$ and $\beta, \gamma, \alpha$ are in geometric progression, then the extreme value of the expression $\gamma x^2 + 4\beta x + \alpha$ is:

• A. $\frac{3}{4}$
• B. $3$
• C. $\frac{3}{2}$
• D. $\frac{21}{4}$ \bigskip

Hint:

In problems involving geometric progressions, use the relationships between the terms to express the roots in terms of a single variable, simplifying calculations.

Answer 1

The Correct Option is C

We are given the cubic equation: $$8x^3 - 42x^2 + 63x - 27 = 0$$ with roots $\alpha, \beta, \gamma$ such that $\beta<\gamma<\alpha$ and they form a geometric progression. 

Step 1: Relationship Between Roots Since $\beta, \gamma, \alpha$ are in geometric progression, we let: $$\beta = \frac{a}{r}, \quad \gamma = a, \quad \alpha = ar$$ Using Vieta’s formulas: 1. Sum of roots: $$\alpha + \beta + \gamma = \frac{42}{8} = \frac{21}{4}$$ Substituting values: $$ar + \frac{a}{r} + a = \frac{21}{4}$$ Multiplying both sides by $r$: $$a r^2 + a + a r = \frac{21}{4}r$$ Factoring: $$a(r^2 + r + 1) = \frac{21}{4}r$$ 2. Product of roots: $$\alpha \beta \gamma = \frac{27}{8}$$ Substituting values: $$(ar) \cdot \left(\frac{a}{r}\right) \cdot a = \frac{27}{8}$$ $$a^3 = \frac{27}{8}$$ $$a = \frac{3}{2}$$ 

Step 2: Finding the Extreme Value of $\gamma x^2 + 4\beta x + \alpha$ The given quadratic expression: $$f(x) = \gamma x^2 + 4\beta x + \alpha$$ Since the coefficient of $x^2$ is $\gamma$, the extreme value occurs at: $$x = -\frac{4\beta}{2\gamma} = -\frac{2\beta}{\gamma}$$ Substituting $\beta = \frac{a}{r}$ and $\gamma = a$: $$x = -\frac{2(a/r)}{a} = -\frac{2}{r}$$ The extreme value is: $$f\left(-\frac{2}{r}\right) = \gamma \left(-\frac{2}{r}\right)^2 + 4\beta \left(-\frac{2}{r}\right) + \alpha$$ $$= a \left(\frac{4}{r^2}\right) - 8 \frac{a}{r} + ar$$ $$= \frac{4a}{r^2} - \frac{8a}{r} + ar$$ Substituting $a = \frac{3}{2}$ and solving, we get: $$\frac{3}{2}$$ 

Final Answer: $\boxed{\frac{3}{2}}$