Question

If the roots of the equation $ax^3+bx^2-x+1=0$ are in H.P. then $\left|\frac{b_{max}}{a_{min}}\right|$ equals ____.

Answer 1

9

Answer 2

Solution:

Let the roots of

$$ax^3 + bx^2 - x + 1 = 0$$

be $\alpha$, $\beta$, and $\gamma$ in harmonic progression (H.P.). Then, the reciprocals $\frac{1}{\alpha}$, $\frac{1}{\beta}$, $\frac{1}{\gamma}$ are in arithmetic progression (A.P.), so

$$\frac{1}{\beta} = \frac{1}{2}\left(\frac{1}{\alpha} + \frac{1}{\gamma}\right) \quad \Longrightarrow \quad 2\alpha\gamma = \beta(\alpha+\gamma).$$

From Vieta’s formulas for the cubic $ax^3+bx^2- x +1 = 0$:

1. $\alpha + \beta + \gamma = -\frac{b}{a}$,
2. $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{-1}{a}$,
3. $\alpha\beta\gamma = -\frac{1}{a}$.

Since $\alpha\beta\gamma = -\frac{1}{a}$, we have

$$\alpha\gamma = \frac{-1}{a\beta}.$$

Also, from the sum we write:

$$\alpha + \gamma = \alpha+\beta+\gamma -\beta = -\frac{b}{a} -\beta.$$

Substitute into the H.P. condition:

$$2\left(\frac{-1}{a\beta}\right) = \beta\left(-\frac{b}{a}-\beta\right).$$

Multiply both sides by $a\beta$ (noting $\beta \neq 0$):

$$-2 = a\beta^2\left(-\frac{b}{a} - \beta\right) = -b\beta^2 - a\beta^3.$$

Rearrange:

$$a\beta^3 + b\beta^2 - 2 = 0. \quad (1)$$

But $\beta$ is a root of the original cubic, so it satisfies:

$$a\beta^3 + b\beta^2 - \beta + 1 = 0. \quad (2)$$

Subtract (1) from (2):

$$\left[a\beta^3 + b\beta^2 - \beta + 1\right] - \left[a\beta^3 + b\beta^2 - 2\right] = -\beta + 3 = 0,$$

which gives

$$\beta = 3.$$

Now substitute $\beta = 3$ in equation (1):

$$a(27) + b(9) - 2= 0 \quad \Longrightarrow \quad 27a + 9b = 2 \quad \Longrightarrow \quad b = \frac{2-27a}{9}.$$

For the roots (other than $\beta=3$) to be real and distinct, consider the quadratic whose roots are $\alpha$ and $\gamma$. From Vieta for these two roots, using:

$$\alpha + \gamma = -\frac{b}{a} - 3,\quad \alpha\gamma = \frac{-1}{3a}.$$

Their discriminant is:

$$\Delta = (\alpha+\gamma)^2 - 4\alpha\gamma = \left(-\frac{2}{9a}\right)^2 - 4\left(-\frac{1}{3a}\right) = \frac{4}{81a^2} + \frac{4}{3a} = \frac{4+108a}{81a^2}.$$

For $\Delta>0$, we require:

$$4 + 108a > 0 \quad \Longrightarrow \quad a > -\frac{1}{27}.$$

The ratio $\left|\frac{b_{\max}}{a_{\min}}\right|$ is achieved when $b$ is maximum and $a$ is minimum. The expression $b = \frac{2}{9} - 3a$ shows that $b$ is maximum when $a$ is minimum (i.e. as low as possible). From the condition, the minimum possible $a$ is just above $-\frac{1}{27}$; taking the infimum, we set:

$$a_{\min} = -\frac{1}{27}.$$

Then,

$$b_{\max} = \frac{2-27\left(-\frac{1}{27}\right)}{9} = \frac{2+1}{9} = \frac{3}{9} = \frac{1}{3}.$$

Thus,

$$\left|\frac{b_{\max}}{a_{\min}}\right| = \left|\frac{\frac{1}{3}}{-\frac{1}{27}}\right| = \left|\frac{1}{3} \times \frac{27}{1}\right| = 9.$$

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Minimal Explanation:

1. For roots in H.P., reciprocals are in A.P. ⇒ $2\alpha\gamma = \beta(\alpha+\gamma)$.
2. Using Vieta’s formulas and eliminating $\alpha$, $\gamma$ yields $\beta=3$ and $27a+9b-2=0$.
3. Express $b$ as $b=\frac{2}{9}-3a$. For real distinct $\alpha$ and $\gamma$, $a > -\frac{1}{27}$.
4. Minimum $a$ is $-\frac{1}{27}$ giving maximum $b=\frac{1}{3}$.
5. Hence, $\left|\frac{b_{\max}}{a_{\min}}\right| = 9$.