prmo 2013 question 16

4. Let 

$f(x) = x^3 - 3x + b$ and $g(x) = x^2 + bx - 3$, where $b$ is a real number. What is the sum of all possible value of $b$ for which the equation $f(x) = 0$ and $g(x) = 0$ have a common root?
[PRE-RMO – 2013]

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Solution:

Suppose 

$x=\alpha$ is a common root of

$$f(x) \;=\; x^{3}-3x+b \quad\text{and}\quad g(x) \;=\; x^{2}+bx-3.$$

Then

$$\alpha^{3}-3\alpha + b \;=\; 0,$$ $$\alpha^{2}+b\,\alpha - 3 \;=\; 0.$$

From the second equation, provided $\alpha \neq 0$, solve for $b$:

$$b \;=\;\frac{3 - \alpha^{2}}{\alpha}.$$

Substitute this into the first equation:

$$\alpha^{3} - 3\alpha \;+\; \frac{3 - \alpha^{2}}{\alpha} \;=\; 0,$$

and multiply through by $\alpha$ (again using $\alpha \neq 0$):

$$\alpha^{4} \;-\; 3\alpha^{2} \;+\;(3 \;-\;\alpha^{2}) \;=\; \alpha^{4} \;-\; 4\alpha^{2} \;+\; 3 \;=\; 0.$$

Make the substitution $y=\alpha^{2}$. Then

$$y^{2} - 4y + 3 \;=\; 0 \quad\Longrightarrow\quad (y-3)(y-1)=0 \quad\Longrightarrow\quad y=1\;\text{or}\;y=3.$$

Hence $\alpha^2=1$ (so $\alpha=\pm 1$) or $\alpha^2=3$ (so $\alpha=\pm\sqrt{3}$). In each case,

$$b \;=\; \frac{3-\alpha^2}{\alpha}$$

gives the following possibilities:

$$\begin{aligned} \alpha = 1 &\quad\Longrightarrow\quad b = 2,\\ \alpha = -1 &\quad\Longrightarrow\quad b = -2,\\ \alpha = \pm\sqrt{3} &\quad\Longrightarrow\quad b = 0. \end{aligned}$$

Thus the values of $b$ that yield a common root are $2,\,-2,$ and $0$. Their sum is

$$2+(-2)+0=0.$$