Key Concepts and Formulas

• Absolute Value Definition: $|x| = x$ if $x \ge 0$, and $|x| = -x$ if $x < 0$.
• Quadratic Formula: For a quadratic equation $ax^2 + bx + c = 0$, the solutions are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Discriminant: For a quadratic equation $ax^2 + bx + c = 0$, the discriminant is $D = b^2 - 4ac$. If $D < 0$, the equation has no real roots.

Step 1: Simplify the Equation with Substitution

We are given the equation $3x(3x - 1) + 2 = |3x - 1| + |3x - 2|$ To simplify the equation, we substitute $t = 3x$. This makes the equation easier to manipulate and understand. Since $x$ is a real number, $t$ is also a real number. Substituting $t$ into the equation, we get $t(t - 1) + 2 = |t - 1| + |t - 2|$ Expanding the left side gives $t^2 - t + 2 = |t - 1| + |t - 2|$

Step 2: Identify Critical Points

The critical points are the values of $t$ where the expressions inside the absolute value signs are equal to zero.

• $t - 1 = 0 \implies t = 1$
• $t - 2 = 0 \implies t = 2$ These critical points divide the real number line into three intervals: $t < 1$, $1 \le t < 2$, and $t \ge 2$. We will analyze each interval separately.

Step 3: Case 1: $t < 1$

If $t < 1$, then $t - 1 < 0$ and $t - 2 < 0$. Therefore, $|t - 1| = -(t - 1) = 1 - t$ and $|t - 2| = -(t - 2) = 2 - t$. Substituting these into the equation, we get $t^2 - t + 2 = (1 - t) + (2 - t)$ $t^2 - t + 2 = 3 - 2t$ Rearranging the terms, we have $t^2 + t - 1 = 0$ Using the quadratic formula, $t = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$ We have two possible solutions: $t_1 = \frac{-1 + \sqrt{5}}{2} \approx 0.618$ and $t_2 = \frac{-1 - \sqrt{5}}{2} \approx -1.618$. Since both $t_1 < 1$ and $t_2 < 1$, both solutions are valid in this interval.

Step 4: Case 2: $1 \le t < 2$

If $1 \le t < 2$, then $t - 1 \ge 0$ and $t - 2 < 0$. Therefore, $|t - 1| = t - 1$ and $|t - 2| = -(t - 2) = 2 - t$. Substituting these into the equation, we get $t^2 - t + 2 = (t - 1) + (2 - t)$ $t^2 - t + 2 = 1$ Rearranging the terms, we have $t^2 - t + 1 = 0$ The discriminant is $D = (-1)^2 - 4(1)(1) = 1 - 4 = -3 < 0$. Therefore, there are no real solutions in this interval.

Step 5: Case 3: $t \ge 2$

If $t \ge 2$, then $t - 1 > 0$ and $t - 2 \ge 0$. Therefore, $|t - 1| = t - 1$ and $|t - 2| = t - 2$. Substituting these into the equation, we get $t^2 - t + 2 = (t - 1) + (t - 2)$ $t^2 - t + 2 = 2t - 3$ Rearranging the terms, we have $t^2 - 3t + 5 = 0$ The discriminant is $D = (-3)^2 - 4(1)(5) = 9 - 20 = -11 < 0$. Therefore, there are no real solutions in this interval.

Step 6: Back-Substitution and Find Roots for x

From Case 1, we found two valid solutions for $t$: $t_1 = \frac{-1 + \sqrt{5}}{2}$ and $t_2 = \frac{-1 - \sqrt{5}}{2}$. Since $t = 3x$, we have $x = \frac{t}{3}$. Thus, $x_1 = \frac{-1 + \sqrt{5}}{6}$ $x_2 = \frac{-1 - \sqrt{5}}{6}$ These are the two distinct real roots for $x$.

Step 7: Form the Set of Real Roots (S) and Final Conclusion

The set $S$ of real roots is $S = \left\{ \frac{-1 + \sqrt{5}}{6}, \frac{-1 - \sqrt{5}}{6} \right\}$. This set contains exactly two elements.

Therefore, the set $S$ contains exactly two elements.

Common Mistakes & Tips

• Remember to check if the solutions obtained in each case satisfy the initial condition for that case.
• Be careful with the signs when dealing with absolute values.
• Don't forget to substitute back to the original variable to find the roots for $x$.

Summary

We solved the equation $3x(3x - 1) + 2 = |3x - 1| + |3x - 2|$ by substituting $t = 3x$, identifying critical points, and analyzing the equation in each interval. We found two real solutions for $t$ in the interval $t < 1$, and no real solutions in the other intervals. Substituting back to $x$, we found two distinct real roots for $x$. Therefore, the set $S$ of real roots contains exactly two elements.

The final answer is \boxed{contains exactly two elements}, which corresponds to option (A).