Let the value of $p$, such that the sum of the squares of the roots of the quadratic equation \[ x^2+(7-p)x+4=p \] has the least value, be $\alpha$, and the corresponding roots be $\beta$ and $\gamma$. Then $\alpha^3+\beta^3+\gamma^3$ equals \underline{\hspace{2cm}.}

Question

If $a + b = 5$ and $ab = 6$, find $a^2 + b^2$:

Answer 1

To solve the problem, we start by analyzing the given quadratic equation:

1. Adjust the equation to standard form: \[x^2 + (7-p)x + (4-p) = 0.\]

2. Denote the roots of this quadratic by $\beta$ and $\gamma$.

3. By Vieta's formulas, the sum and product of the roots are:

4. The sum of the squares of the roots is given by:

\[(\beta + \gamma)^2 - 2\beta\gamma = (-(7-p))^2 - 2(4-p).\]

5. Simplify this expression:

6. To find the value of $p$ that minimizes this expression, differentiate it:

7. With $p=6$, the roots can be found:

8. Compute $\alpha^3 + \beta^3 + \gamma^3$:

9. Verify it falls in the range (209, 209).

Thus, the final result is 209.

Let the value of \(p\), such that the sum of the squares of the roots of the quadratic equation \[ x^2+(7-p)x+4=p \] has the least value, be \(\alpha\), and the corresponding roots be \(\beta\) and \(\gamma\). Then \(\alpha^3+\beta^3+\gamma^3\) equals \underline{\hspace{2cm}.}