How many solutions exist for the equation 3x^10 = 6 + 3x^12?

To determine how many solutions exist for the equation 3x¹⁰ = 6 + 3x¹², we first need to rearrange it into a standard form where one side equals zero.

Start by subtracting 3x¹² from both sides:

3x10 - 3x12 - 6 = 0

Next, factor out 3 from the left-hand side:

3(x10 - x12 - 2) = 0

This simplifies to:

x10 - x12 - 2 = 0

Now, let’s simplify the expression further:

x10(1 - x2) - 2 = 0

From here, we can set this whole equation to find the critical points:

x10(1 - x2) = 2

Analyze this equation:

• x¹⁰ is always non-negative since any base raised to an even power is positive or zero.
• 1 – x² becomes zero when x = ±1. It will be positive when -1 < x < 1 and negative otherwise.

Now, let’s check some possible values:

• If -1 < x < 1, then the term 1 – x² is positive. Therefore, the entire expression can potentially equal 2 depending on the value of x¹⁰.
• If x > 1 or x < -1, then 1 – x² becomes negative, leading to a negative product that cannot equal 2.

This means that solutions will only exist when -1 < x < 1. To find the exact number of solutions:

• You would apply numerical methods (like Newton’s method) or graphical analysis to find the intersection of the curves formed by x¹⁰(1 – x²) and 2.
• Given that the function is continuous within this interval and based on the intermediate value theorem, there should be at least two solutions within -1 < x < 1, as it will cross the line y = 2.

In conclusion, the number of solutions to the equation 3x¹⁰ = 6 + 3x¹² is likely two, found between the intervals of -1 and 1.