$$ \left(10 x^{2}-9\right)^{\frac{1}{4}}=x $$ What is the sum of the solutions to the given equation?

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To solve the equation $$ \left(10 x^{2}-9\right)^{\frac{1}{4}} = x $$, follow these steps:

1. **Raise Both Sides to the Fourth Power**:
   $$
   10x^2 - 9 = x^4
   $$
   
2. **Rearrange the Equation**:
   $$
   x^4 - 10x^2 + 9 = 0
   $$
   
3. **Solve the Biquadratic Equation**:
   Let $$ y = x^2 $$, the equation becomes:
   $$
   y^2 - 10y + 9 = 0
   $$
   
   Using the quadratic formula:
   $$
   y = \frac{10 \pm \sqrt{100 - 36}}{2} = \frac{10 \pm 8}{2}
   $$
   Thus, $$ y = 9 $$ or $$ y = 1 $$.

4. **Find $$ x $$ from $$ y $$**:
   - If $$ y = 9 $$, then $$ x = \pm 3 $$.
   - If $$ y = 1 $$, then $$ x = \pm 1 $$.
   
5. **Check for Valid Solutions in the Original Equation**:
   - **$$ x = 3 $$**:
     $$
     \left(10(3)^2 - 9\right)^{1/4} = \left(90 - 9\right)^{1/4} = 81^{1/4} = 3 \quad \text{(Valid)}
     $$
     
   - **$$ x = -3 $$**:
     $$
     \left(10(3)^2 - 9\right)^{1/4} = 81^{1/4} = 3 \neq -3 \quad \text{(Invalid)}
     $$
     
   - **$$ x = 1 $$**:
     $$
     \left(10(1)^2 - 9\right)^{1/4} = 1^{1/4} = 1 \quad \text{(Valid)}
     $$
     
   - **$$ x = -1 $$**:
     $$
     \left(10(1)^2 - 9\right)^{1/4} = 1^{1/4} = 1 \neq -1 \quad \text{(Invalid)}
     $$
   
   Therefore, the valid solutions are $$ x = 3 $$ and $$ x = 1 $$.

6. **Sum of the Solutions**:
   $$
   3 + 1 = 4
   $$

**Answer:** 4
The sum of the solutions is 4.

\( \left(10 x^{2}-9\right)^{\frac{1}{4}}=x \) What is the sum of the solutions to the given equation?

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