Key Concepts and Formulas

• The fundamental identity: $x^2 = |x|^2$
• Solving quadratic equations of the form $ax^2 + bx + c = 0$ by factoring or using the quadratic formula.
• Definition of absolute value: If $|x| = k$ where $k > 0$, then $x = k$ or $x = -k$.

Step-by-Step Solution

Step 1: Transform the given equation using the identity $x^2 = |x|^2$

Given the equation: $9x^2 - 18|x| + 5 = 0$

We know that $x^2 = |x|^2$. Substituting this into the equation, we get: $9|x|^2 - 18|x| + 5 = 0$

Explanation: By replacing $x^2$ with $|x|^2$, we transform the original equation into a quadratic equation in terms of $|x|$. This allows us to solve the equation using standard quadratic techniques.

Step 2: Solve the quadratic equation for $|x|$

Let $y = |x|$. The equation becomes a standard quadratic equation in $y$: $9y^2 - 18y + 5 = 0$

We can solve this quadratic equation by factorization. We need two numbers whose product is $9 \times 5 = 45$ and whose sum is $-18$. These numbers are $-15$ and $-3$.

Rewrite the middle term: $9y^2 - 15y - 3y + 5 = 0$

Factor by grouping: $3y(3y - 5) - 1(3y - 5) = 0$

Factor out the common binomial $(3y - 5)$: $(3y - 5)(3y - 1) = 0$

Now, set each factor to zero to find the values of $y$: $3y - 5 = 0 \quad \Rightarrow \quad 3y = 5 \quad \Rightarrow \quad y = \frac{5}{3}$ $3y - 1 = 0 \quad \Rightarrow \quad 3y = 1 \quad \Rightarrow \quad y = \frac{1}{3}$

Explanation: We solved the quadratic equation for $y$, which represents $|x|$. These values are the possible magnitudes for $x$.

Step 3: Determine the real roots for $x$ from the values of $|x|$

Since $y = |x|$, we have two cases:

Case 1: $|x| = \frac{5}{3}$ The definition of absolute value states that if $|x| = k$ (where $k > 0$), then $x = k$ or $x = -k$. Therefore, for $|x| = \frac{5}{3}$, the roots are: $x = \frac{5}{3} \quad \text{or} \quad x = -\frac{5}{3}$

Case 2: $|x| = \frac{1}{3}$ Similarly, for $|x| = \frac{1}{3}$, the roots are: $x = \frac{1}{3} \quad \text{or} \quad x = -\frac{1}{3}$

Thus, the four real roots of the original equation are $x = \frac{1}{3}, -\frac{1}{3}, \frac{5}{3}, -\frac{5}{3}$.

Explanation: Each positive value obtained for $|x|$ yields two distinct real roots for $x$ because a number and its negative both have the same absolute value. It is crucial to consider both the positive and negative counterparts.

Step 4: Calculate the product of all roots

Now, we multiply all the roots found in Step 3: Product of roots $= \left(\frac{1}{3}\right) \times \left(-\frac{1}{3}\right) \times \left(\frac{5}{3}\right) \times \left(-\frac{5}{3}\right)$ $= \left(\frac{1}{3} \times \frac{-1}{3}\right) \times \left(\frac{5}{3} \times \frac{-5}{3}\right)$ $= \left(-\frac{1}{9}\right) \times \left(-\frac{25}{9}\right)$ $= \frac{(-1) \times (-25)}{9 \times 9}$ $= \frac{25}{81}$

Explanation: We multiply all four roots together. Remember that the product of an even number of negative terms results in a positive product.

Common Mistakes & Tips

• Don't forget the negative roots: When solving $|x| = k$, remember that $x$ can be both $k$ and $-k$.
• Check for extraneous solutions: Ensure that the values of $|x|$ you obtain are non-negative. If $|x|$ is negative, there are no real solutions for that case.
• Careful with signs: Double-check the signs when multiplying the roots to ensure you get the correct final answer.

Summary

To solve equations involving both $x^2$ and $|x|$, leverage the identity $x^2 = |x|^2$ to convert the equation into a quadratic form in terms of $|x|$. Solve this quadratic equation for $|x|$, and then for each valid (non-negative) solution of $|x|$, find the corresponding two real roots for $x$ (a positive and a negative value). Finally, multiply all such roots to get the required product. The product of the roots for the given equation is $\frac{25}{81}$.

Final Answer

The final answer is \boxed{\frac{25}{81}}, which corresponds to option (C).