Given that the three points where the curve `y=bx^(2)-2` intersects the x-axis and y-axis form an equilateral triangle. Find the value of 2b.

Answer

To solve the problem, we need to find the value of \(2b\) given that the points where the curve \(y = bx^2 - 2\) intersects the x-axis and y-axis form an equilateral triangle.

### Step-by-step Solution:

1. **Find the intersection points with the x-axis**:
The curve intersects the x-axis when \(y = 0\).
\[ 0 = bx^2 - 2 \]
Rearranging gives:
\[ bx^2 = 2 \quad \Rightarrow \quad x^2 = \frac{2}{b} \]
Thus, the x-coordinates of the intersection points are:
\[ x = \pm \sqrt{\frac{2}{b}} \]
Let’s denote these points as \(A\) and \(B\):
\[ A\left(\sqrt{\frac{2}{b}}, 0\right), \quad B\left(-\sqrt{\frac{2}{b}}, 0\right) \]

2. **Find the intersection point with the y-axis**:
The curve intersects the y-axis when \(x = 0\).
\[ y = b(0)^2 - 2 = -2 \]
Thus, the intersection point is:
\[ C(0, -2) \]

3. **Determine the lengths of the sides of the triangle**:
The distance \(AB\) between points \(A\) and \(B\) is:
\[ AB = \sqrt{\left(\sqrt{\frac{2}{b}} - (-\sqrt{\frac{2}{b}})\right)^2 + (0 - 0)^2} = \sqrt{\left(2\sqrt{\frac{2}{b}}\right)^2} = 2\sqrt{\frac{2}{b}} = \frac{2\sqrt{2}}{\sqrt{b}} \]

The distance \(AC\) between points \(A\) and \(C\) is:
\[ AC = \sqrt{\left(\sqrt{\frac{2}{b}} - 0\right)^2 + (0 - (-2))^2} = \sqrt{\left(\sqrt{\frac{2}{b}}\right)^2 + 2^2} = \sqrt{\frac{2}{b} + 4} \]

The distance \(BC\) between points \(B\) and \(C\) is:
\[ BC = \sqrt{\left(-\sqrt{\frac{2}{b}} - 0\right)^2 + (0 - (-2))^2} = \sqrt{\left(-\sqrt{\frac{2}{b}}\right)^2 + 2^2} = \sqrt{\frac{2}{b} + 4} \]

4. **Set the distances equal for an equilateral triangle**:
Since \(AB = AC = BC\), we have:
\[ AB = AC \]
Therefore,
\[ \frac{2\sqrt{2}}{\sqrt{b}} = \sqrt{\frac{2}{b} + 4} \]
Squaring both sides:
\[ \left(\frac{2\sqrt{2}}{\sqrt{b}}\right)^2 = \frac{2}{b} + 4 \]
Simplifying gives:
\[ \frac{8}{b} = \frac{2}{b} + 4 \]
Multiplying through by \(b\) (assuming \(b \neq 0\)):
\[ 8 = 2 + 4b \]
Rearranging gives:
\[ 6 = 4b \quad \Rightarrow \quad b = \frac{3}{2} \]

5. **Calculate \(2b\)**:
\[ 2b = 2 \times \frac{3}{2} = 3 \]

### Final Answer:
\[ \boxed{3} \]