solve x^2+(9/x^2)=10 for x such that x also satisfies (x^3)-3x>0

We are tasked with solving \( x^2 + \frac{9}{x^2} = 10 \) for \( x \), and then determining which solutions satisfy \( x^3 - 3x > 0 \).

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

Let \( y = x^2 \). This transforms the equation to:
\[
y + \frac{9}{y} = 10
\]
Multiply through by \( y \) to eliminate the fraction:
\[
y^2 + 9 = 10y
\]
Rearrange the terms to form a quadratic equation:
\[
y^2 - 10y + 9 = 0
\]
Solve this quadratic using the quadratic formula:
\[
y = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(9)}}{2(1)}
\]
\[
y = \frac{10 \pm \sqrt{100 - 36}}{2}
\]
\[
y = \frac{10 \pm \sqrt{64}}{2}
\]
\[
y = \frac{10 \pm 8}{2}
\]
This gives two solutions:
\[
y = \frac{10 + 8}{2} = 9 \quad \text{or} \quad y = \frac{10 - 8}{2} = 1
\]

### Step 2: Solve for \( x \)

Since \( y = x^2 \), we have two cases:
1. \( x^2 = 9 \), so \( x = \pm 3 \)
2. \( x^2 = 1 \), so \( x = \pm 1 \)

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

### Step 3: Check the inequality \( x^3 - 3x > 0 \)

We now check which of these values satisfy the inequality \( x^3 - 3x > 0 \).

1. For \( x = 3 \):
   \[
   x^3 - 3x = 3^3 - 3(3) = 27 - 9 = 18 > 0
   \]
   So, \( x = 3 \) satisfies the inequality.

2. For \( x = -3 \):
   \[
   x^3 - 3x = (-3)^3 - 3(-3) = -27 + 9 = -18 < 0
   \]
   So, \( x = -3 \) does not satisfy the inequality.

3. For \( x = 1 \):
   \[
   x^3 - 3x = 1^3 - 3(1) = 1 - 3 = -2 < 0
   \]
   So, \( x = 1 \) does not satisfy the inequality.

4. For \( x = -1 \):
   \[
   x^3 - 3x = (-1)^3 - 3(-1) = -1 + 3 = 2 > 0
   \]
   So, \( x = -1 \) satisfies the inequality.

### Final Solution

The values of \( x \) that satisfy both the equation and the inequality are:
\[
x = 3 \quad \text{or} \quad x = -1
\]

Would you like further details or have any questions?

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**Related Questions:**
1. What would happen if \( x^2 + \frac{9}{x^2} = 15 \)?
2. How does the inequality \( x^3 - 3x < 0 \) change the solution set?
3. Can we solve \( x^2 + \frac{k}{x^2} = n \) for any \( k \) and \( n \)?
4. What happens if we replace \( 9/x^2 \) with \( 9/x^3 \)?
5. How would the solution change if the inequality were \( x^3 - 2x > 0 \)?

**Tip:** Always check for extraneous solutions when solving equations involving fractions and higher powers.

Explore the step-by-step solution to the equation x^2 + (9/x^2) = 10 and check which values of x satisfy the inequality x^3 - 3x > 0. This problem involves quadratic equations, inequalities, and applying algebraic concepts suitable for Grades 10-12.

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