Consider the equation x²+(9)/(x^(2))= 10. Solve for x such that the inequality x³-3x > 0 is also satisfied. Additionally, show that y = x³-3x is increasing at x = 1.

Let's break the problem into two parts:

### Part 1: Solve the equation \(x^2 + \frac{9}{x^2} = 10\)

First, multiply both sides of the equation by \(x^2\) to eliminate the fraction:

\[
x^4 + 9 = 10x^2
\]

Rearrange the terms to form a quadratic in \(x^2\):

\[
x^4 - 10x^2 + 9 = 0
\]

Now, let \(z = x^2\). The equation becomes:

\[
z^2 - 10z + 9 = 0
\]

Solve this quadratic equation using the quadratic formula:

\[
z = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(9)}}{2(1)} = \frac{10 \pm \sqrt{100 - 36}}{2} = \frac{10 \pm \sqrt{64}}{2}
\]

\[
z = \frac{10 \pm 8}{2}
\]

Thus, we get two possible values for \(z\):

\[
z = \frac{10 + 8}{2} = 9 \quad \text{or} \quad z = \frac{10 - 8}{2} = 1
\]

Since \(z = x^2\), we now have two possibilities:

1. \(x^2 = 9 \implies x = \pm 3\)
2. \(x^2 = 1 \implies x = \pm 1\)

So, the possible values of \(x\) are \(x = -3, -1, 1, 3\).

### Part 2: Solve the inequality \(x^3 - 3x > 0\)

Factor the inequality:

\[
x(x^2 - 3) > 0
\]

This factors into:

\[
x(x - \sqrt{3})(x + \sqrt{3}) > 0
\]

To solve this inequality, we analyze the sign of \(x(x - \sqrt{3})(x + \sqrt{3})\) over different intervals by considering the critical points \(x = 0, x = \sqrt{3}, x = -\sqrt{3}\). The sign chart for these intervals is:

- For \(x < -\sqrt{3}\), the expression is negative.
- For \(-\sqrt{3} < x < 0\), the expression is positive.
- For \(0 < x < \sqrt{3}\), the expression is negative.
- For \(x > \sqrt{3}\), the expression is positive.

Thus, the inequality holds for \(x \in (-\sqrt{3}, 0) \cup (\sqrt{3}, \infty)\).

### Combine both conditions

The possible solutions from the equation are \(x = -3, -1, 1, 3\). Now, we apply the inequality condition \(x \in (-\sqrt{3}, 0) \cup (\sqrt{3}, \infty)\):

- \(x = -3\) does not satisfy the inequality.
- \(x = -1\) does not satisfy the inequality.
- \(x = 1\) does not satisfy the inequality.
- \(x = 3\) satisfies the inequality.

Thus, the only solution is \(x = 3\).

### Part 3: Show that \(y = x^3 - 3x\) is increasing at \(x = 1\)

To determine whether \(y = x^3 - 3x\) is increasing at \(x = 1\), we calculate the derivative of \(y\) with respect to \(x\):

\[
\frac{dy}{dx} = 3x^2 - 3
\]

Now, evaluate the derivative at \(x = 1\):

\[
\frac{dy}{dx}\bigg|_{x=1} = 3(1)^2 - 3 = 3 - 3 = 0
\]

Since the derivative is zero at \(x = 1\), \(y = x^3 - 3x\) is neither increasing nor decreasing at \(x = 1\). Instead, there is a critical point at \(x = 1\), suggesting a possible local minimum or maximum, but it is not strictly increasing.

---

Would you like more details or clarification on any step? Here are 5 related questions to expand on this concept:

1. How do we analyze the nature of critical points for a cubic function like \(y = x^3 - 3x\)?
2. Can you explain how to solve inequalities involving factored cubic expressions?
3. What is the geometric interpretation of solving inequalities like \(x^3 - 3x > 0\)?
4. How can we verify if \(x = 1\) is a local minimum or maximum for \(y = x^3 - 3x\)?
5. How would the solution set change if the equation was \(x^2 + \frac{9}{x^2} = 8\)?

**Tip:** For cubic inequalities, always factor and check sign changes across critical points to find solution intervals.

Consider the equation x² + (9)/(x²) = 10. Solve for x such that the inequality x³ - 3x > 0 is also satisfied. Additionally, show that y = x³ - 3x is increasing at x = 1.

Learn how to solve the equation x² + 9/x² = 10 and apply conditions to satisfy the inequality x³ - 3x > 0. The problem covers algebraic techniques, solving quadratic equations, and analyzing inequalities. Suitable for students in Grades 10-12.

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