Question:

The roots of \(x^2 – 5x + 3 = 0\) are \(\alpha\) and \(\beta\). Another quadratic equation \(x^2 + px + q =0\) has roots, \(\alpha^3\) and \(\beta^3\). Then, find \(p\) and \(q\).

Answer:

If \(\alpha\) and \(\beta\) are the roots of a quadratic equation, \(ax^2 + bx + c = 0 \quad (a\neq 0)\), the coefficients and roots have the following relationship:

\[  \alpha + \beta = -\frac{b}{a}, \]

\[  \alpha \cdot \beta = \frac{c}{a}  \]

From the first equation, we have

\[  \alpha + \beta = 5, \]

\[  \alpha \cdot \beta = 3  \]

From the other equation, they can be expressed as

\[  \alpha^3+\beta^3 = -p, \]

\[  (\alpha\beta)^3 = q  \]

Thus, we can easily get

\[ q = 27 \]

However, \(\alpha^3+\beta^3\) should be factored out as follows:

\[  \alpha^3+\beta^3 = (\alpha+\beta)(\alpha^2 -\alpha\beta + \beta^2)  \]

\[                             = (\alpha+\beta)(\alpha^2 +2\alpha\beta + \beta^2 – 3\alpha\beta)  \]

\[                            = (\alpha+\beta)((\alpha + \beta)^2 – 3\alpha\beta)  \]

Therefore, we obtain

\[  p = – (\alpha+\beta)((\alpha + \beta)^2 – 3\alpha\beta) = 5 \cdot (5^2 – 3 \cdot 3) = -80  \]