If the sum of the roots of an equation is 2 and the sum of their cubes is 98, then find the equation.

Explanation

To find the equation given the sum of the roots and the sum of their cubes, we can follow these steps:

**Step 1: Define the roots.**
Let the roots of the equation be \( \alpha \) and \( \beta \).

**Step 2: Use the given information.**
According to the problem, we have:
- The sum of the roots:
\[ \alpha + \beta = 2 \]
- The sum of the cubes of the roots:
\[ \alpha^3 + \beta^3 = 98 \]

**Step 3: Use the identity for the sum of cubes.**
We can use the identity for the sum of cubes:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) \]
We can also express \( \alpha^2 + \beta^2 \) in terms of \( \alpha + \beta \) and \( \alpha\beta \):
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]

**Step 4: Substitute known values.**
Substituting \( \alpha + \beta = 2 \) into the identity gives:
\[ \alpha^3 + \beta^3 = 2(\alpha^2 - \alpha\beta + \beta^2) \]
Now substituting \( \alpha^2 + \beta^2 \):
\[ \alpha^3 + \beta^3 = 2\left((\alpha + \beta)^2 - 3\alpha\beta\right) \]
Substituting \( \alpha + \beta = 2 \):
\[ \alpha^3 + \beta^3 = 2\left(2^2 - 3\alpha\beta\right) = 2(4 - 3\alpha\beta) = 8 - 6\alpha\beta \]

**Step 5: Set up the equation.**
Now we know that:
\[ 8 - 6\alpha\beta = 98 \]
Rearranging gives:
\[ -6\alpha\beta = 98 - 8 \]
\[ -6\alpha\beta = 90 \]
Dividing both sides by -6:
\[ \alpha\beta = -15 \]

**Step 6: Form the quadratic equation.**
Now we have both the sum and product of the roots:
- \( \alpha + \beta = 2 \)
- \( \alpha\beta = -15 \)

The quadratic equation can be formed using the standard form:
\[ x^2 - (\alpha + \beta)x + \alpha\beta = 0 \]
Substituting the values:
\[ x^2 - 2x - 15 = 0 \]

**Final Equation:**
Thus, the required equation is:
\[ x^2 - 2x - 15 = 0 \]

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