consider the equation x²+9/x²=10. solve for such that the inequality x³-3x>0 is also satisfied addtionally, show that y=x³-3x is increassing at x=1

We are given two tasks to solve:

1. Solve the equation:
   \[
   x^2 + \frac{9}{x^2} = 10
   \]
   such that the solution also satisfies the inequality:
   \[
   x^3 - 3x > 0
   \]

2. Show that the function \( y = x^3 - 3x \) is increasing at \( x = 1 \).

Let's tackle these one by one.

### Step 1: Solve the Equation \( x^2 + \frac{9}{x^2} = 10 \)
Let \( z = x^2 \). Then \( \frac{9}{x^2} = \frac{9}{z} \), and the equation becomes:
\[
z + \frac{9}{z} = 10
\]
Multiply both sides by \( z \) to eliminate the fraction:
\[
z^2 + 9 = 10z
\]
Rearrange to form a quadratic equation:
\[
z^2 - 10z + 9 = 0
\]
Solve this quadratic equation using the quadratic formula:
\[
z = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 9}}{2 \cdot 1} = \frac{10 \pm \sqrt{100 - 36}}{2} = \frac{10 \pm \sqrt{64}}{2} = \frac{10 \pm 8}{2}
\]
This gives:
\[
z = \frac{18}{2} = 9 \quad \text{or} \quad z = \frac{2}{2} = 1
\]
Since \( z = x^2 \), we have two cases:
1. \( x^2 = 9 \Rightarrow x = \pm 3 \)
2. \( x^2 = 1 \Rightarrow x = \pm 1 \)

Thus, the possible values for \( x \) are \( x = \pm 3 \) and \( x = \pm 1 \).

### Step 2: Satisfy the Inequality \( x^3 - 3x > 0 \)
Let's evaluate the inequality \( x^3 - 3x > 0 \):
\[
x^3 - 3x = x(x^2 - 3)
\]
We can factor \( x^2 - 3 \) as follows:
\[
x(x - \sqrt{3})(x + \sqrt{3}) > 0
\]
This inequality holds for values of \( x \) that are outside the intervals \( (-\sqrt{3}, 0) \) and \( (0, \sqrt{3}) \).

Evaluating each solution:
- \( x = 3 \): This satisfies \( x^3 - 3x > 0 \).
- \( x = -3 \): This does **not** satisfy \( x^3 - 3x > 0 \) since \( -3 \cdot (-3^2 - 3) = -3 \cdot (-12) = 36 \), which is positive.
- \( x = \pm 1 \): When \( x = 1 \), \( y = x^3 - 3x \)

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.

Explore how to solve the equation x² + 9/x² = 10 while satisfying the inequality x³ - 3x > 0. This solution includes factorization, quadratic formula, and derivative analysis to determine when y = x³ - 3x is increasing at x = 1. Suitable for students in Grades 10-12.

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