Key Concepts and Formulas

• Substitution: Identifying repeating expressions in an equation and substituting a new variable to simplify the equation.
• AM-GM Inequality: For non-negative real numbers $a_1, a_2, ..., a_n$, $\frac{a_1 + a_2 + ... + a_n}{n} \ge \sqrt[n]{a_1 a_2 ... a_n}$. In particular, for two positive real numbers $a$ and $b$, $\frac{a+b}{2} \ge \sqrt{ab}$, or $a+b \ge 2\sqrt{ab}$.
• Quadratic Formula/Factoring: Methods to find the roots of a quadratic equation of the form $ax^2 + bx + c = 0$.

Step-by-Step Solution

1. Simplifying the Equation using Substitution

The given equation is: $3\left( {{x^2} + {1 \over {{x^2}}}} \right) - 2\left( {x + {1 \over x}} \right) + 5 = 0$ We notice terms of the form $x + \frac{1}{x}$ and $x^2 + \frac{1}{x^2}$. We want to express $x^2 + \frac{1}{x^2}$ in terms of $x + \frac{1}{x}$. Using the algebraic identity $(a+b)^2 = a^2 + 2ab + b^2$, let $a=x$ and $b=\frac{1}{x}$. Then, $\left(x + \frac{1}{x}\right)^2 = x^2 + 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x^2} = x^2 + 2 + \frac{1}{x^2}$. Rearranging, we get: $x^2 + {1 \over x^2} = \left(x + {1 \over x}\right)^2 - 2$

WHY this step? This allows us to rewrite the given equation in terms of a single expression, $x + \frac{1}{x}$, making it easier to solve.

Let $t = x + \frac{1}{x}$. Then, $x^2 + \frac{1}{x^2} = t^2 - 2$. Substituting these into the original equation: $3(t^2 - 2) - 2(t) + 5 = 0$

2. Determining the Valid Range for the Substituted Variable '$t$'

Before solving for $t$, we need to determine the possible values of $t = x + \frac{1}{x}$ when $x$ is a real number.

WHY this step? Not all solutions for $t$ will yield real solutions for $x$. Restricting the domain of $t$ ensures we only consider solutions that could possibly arise from real values of $x$.

Consider the function $f(x) = x + \frac{1}{x}$ for real $x$:

• If $x > 0$: By the AM-GM inequality, $x + \frac{1}{x} \ge 2\sqrt{x \cdot \frac{1}{x}} = 2$. Equality occurs when $x = \frac{1}{x}$, which means $x^2=1$. Since $x>0$, $x=1$. Thus, for positive $x$, $t \ge 2$.
• If $x < 0$: Let $x = -y$, where $y > 0$. Then $t = x + \frac{1}{x} = -y + \frac{1}{-y} = -\left(y + \frac{1}{y}\right)$. Since $y + \frac{1}{y} \ge 2$ (from the AM-GM case for positive $y$), then $-\left(y + \frac{1}{y}\right) \le -2$. Equality occurs when $y=1$, which means $x=-1$. Thus, for negative $x$, $t \le -2$.

Combining these cases, the valid range for $t$ when $x$ is real is: $t \in (-\infty, -2] \cup [2, \infty)$

3. Solving the Quadratic Equation for '$t$'

Simplify and solve the transformed equation for $t$: $3(t^2 - 2) - 2t + 5 = 0$ $3t^2 - 6 - 2t + 5 = 0$ $3t^2 - 2t - 1 = 0$ We can solve this quadratic equation by factoring: $3t^2 - 3t + t - 1 = 0$ $3t(t - 1) + 1(t - 1) = 0$ $(3t + 1)(t - 1) = 0$ This gives two possible solutions for $t$: $3t + 1 = 0 \implies t = -\frac{1}{3}$ $t - 1 = 0 \implies t = 1$

WHY this step? This step finds the values of $t$ that satisfy the simplified quadratic equation. These values are potential solutions, but they must be validated.

4. Verifying Solutions Against the Valid Range

Compare the values of $t$ we found ($t=1$ and $t=-\frac{1}{3}$) with the valid range for $t$, which is $t \in (-\infty, -2] \cup [2, \infty)$.

• For $t = 1$: This value is not in the valid range because $1$ is not less than or equal to $-2$ and not greater than or equal to $2$.
• For $t = -\frac{1}{3}$: This value is also not in the valid range because $-\frac{1}{3}$ is not less than or equal to $-2$ and not greater than or equal to $2$.

WHY this step? This step is crucial to eliminate extraneous solutions. We are checking if the solutions we found for $t$ are actually possible given the constraint that $x$ must be real.

5. Conclusion: Number of Real Solutions for '$x$'

Since neither of the values obtained for $t$ ($1$ and $-\frac{1}{3}$) falls within the permissible range $(-\infty, -2] \cup [2, \infty)$, there are no real values of $t$ that satisfy the conditions. Consequently, there are no real values of $x$ that can satisfy the original equation.

Therefore, the number of real solutions for the given equation is 0.

Common Mistakes & Tips

• Forgetting the Range of $t$: Always determine the range of $t = x + \frac{1}{x}$ for real $x$ and check your solutions against it.
• Incorrect AM-GM: Make sure you apply AM-GM correctly, especially when dealing with negative values. Consider substituting $x = -y$ when $x < 0$.
• Algebraic Mistakes: Double-check your algebraic manipulations, especially when squaring, expanding, and factoring.

Summary

By substituting $t = x + \frac{1}{x}$ and recognizing that $x^2 + \frac{1}{x^2} = t^2 - 2$, the equation was transformed into a quadratic $3t^2 - 2t - 1 = 0$. The solutions for $t$ were found to be $1$ and $-\frac{1}{3}$. However, for real solutions of $x$, $t$ must lie in the interval $(-\infty, -2] \cup [2, \infty)$. Since neither $1$ nor $-\frac{1}{3}$ falls within this interval, there are no real solutions for $x$.

The final answer is \boxed{0}, which corresponds to option (C).