Quadratic equation: The relationship between solutions and coefficients 2

Question:
The equation, $kx^2 -(k+3)x - 1 =0$, which has real coefficients, has complex roots, $a+ib$ and $a-ib$. In order to have those roots, find the range of $k$. If they are pure imaginary roots, find the value of $k$.

Answer:
The discriminant of the quadratic equation, $ax^2+bx+c=0$, is
\[
D = b^2 - 4ac
\]
For complex roots, we must have $D<0$ and $k \neq 0$. Thus,
\begin{eqnarray*}
D &=& (k+3)^2 +4k    \\
    &=& k^2+10k+9      \\
    &=& (k+1)(k+9)  < 0
\end{eqnarray*}
The range of $k$ is
\[
-9 < k < -1 \quad \mathrm{but} \quad k \neq 0
\]
If the roots are pure imaginary, we let the roots be $ib$ and $-ib$. We have the following relationship:
\[
ib + (-ib) = - \frac{-(k+3)}{k}
\]
This is from the relationship between the roots and coefficients. Then,
\begin{eqnarray*}
0 &=& k +3  \\
\rightarrow k &=& -3
\end{eqnarray*}
This satisfies $-9 < k < -1$.

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Quadratic equation: The relationship between solutions and coefficients 1

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 $\beta^3$. 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
\[
\alpha^3+\beta^3 = -p, \\
(\alpha\beta)^3 = q
\]
We can easily have
\[ q = 27 \]
However, $\alpha^3+\beta^3$ should be modified as follows:
\begin{eqnarray*}
\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)
\end{eqnarray*}
Therefore, we obtain
\[
p = - (\alpha+\beta)((\alpha + \beta)^2 - 3\alpha\beta) = 5 \cdot (5^2 - 3 \cdot 3) = -80
\]

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