If alpha and beta are the roots of the equation 3x2 − 2x − 1 = 0, find the equation whose roots are (alpha)/(beta) and (beta)/(alpha).

For $3x^2 - 2x - 1 = 0$, we have $a = 3$, $b = -2$, $c = -1$.
Sum of roots: $\alpha + \beta = -\frac{b}{a} = -\frac{(-2)}{3} = \frac{2}{3}$.
Product of roots: $\alpha\beta = \frac{c}{a} = \frac{-1}{3}$.

Let the new roots be $r_1 = \frac{\alpha}{\beta}$ and $r_2 = \frac{\beta}{\alpha}$.
Sum of new roots: $S' = \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta}$.
We know that $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$.

Substitute the values of $\alpha + \beta$ and $\alpha\beta$:

Product of new roots: $P' = \left(\frac{\alpha}{\beta}\right) \left(\frac{\beta}{\alpha}\right) = 1$.

The equation is $x^2 - S'x + P' = 0$.

Multiply by 3 to clear the fraction:

The equation is $\boxed{3x^2 + 10x + 3 = 0}$.

6. Without solving the equation, find the sum and product of the roots of $3x^2 + 5x - 2 = 0$.

For $3x^2 + 5x - 2 = 0$, we have $a = 3$, $b = 5$, $c = -2$.

Sum of roots $= -\frac{b}{a} = -\frac{5}{3}$.
The sum of the roots is $\boxed{-\frac{5}{3}}$.

Product of roots $= \frac{c}{a} = \frac{-2}{3}$.
The product of the roots is $\boxed{-\frac{2}{3}}$.

7. The product of two consecutive integers is 132. Find the integers.

Let the first integer be $n$. The next consecutive integer is $n+1$.
Their product is $n(n+1) = 132$.

Factor the quadratic equation:
We need two numbers that multiply to -132 and add to 1. These numbers are 12 and -11.

So, $n = -12$ or $n = 11$.

If $n = 11$, the integers are 11 and $11+1 = 12$.
If $n = -12$, the integers are $-12$ and $-12+1 = -11$.
The integers are $\boxed{11 \text{ and } 12}$ or $\boxed{-12 \text{ and } -11}$.

8. The length of a rectangle exceeds its width by 5 units. If the area is 104 square units, find the dimensions.

Let the width of the rectangle be $w$ units.
The length $l$ exceeds the width by 5 units, so $l = w + 5$.
The area $A$ is 104 square units.
The formula for the area of a rectangle is $A = l \times w$.

Factor the quadratic equation:
We need two numbers that multiply to -104 and add to 5. These numbers are 13 and -8.

So, $w = -13$ or $w = 8$.

Since width cannot be negative, we take $w = 8$ units.
Length $l = w + 5 = 8 + 5 = 13$ units.
The dimensions are $\boxed{\text{width } = 8 \text{ units, length } = 13 \text{ units}}$.

9. Show that if $a < 0$, $y = ax^2 + bx + c$ has a maximum value when $x = -\frac{b}{2a}$.

Factor out $a$ from the terms involving $x$:

To complete the square inside the parenthesis, add and subtract $\left(\frac{b}{2a}\right)^2$:

Distribute $a$:

The term $a\left(x + \frac{b}{2a}\right)^2$ is the only term that depends on $x$.
Given that $a < 0$, the term $a\left(x + \frac{b}{2a}\right)^2$ will always be less than or equal to zero, because $\left(x + \frac{b}{2a}\right)^2$ is always non-negative.
For $y$ to have a maximum value, the term $a\left(x + \frac{b}{2a}\right)^2$ must be at its maximum possible value, which is 0.
This occurs when $\left(x + \frac{b}{2a}\right)^2 = 0$.

When $x = -\frac{b}{2a}$, the value of $y$ is $y = a(0)^2 + \frac{4ac - b^2}{4a} = \frac{4ac - b^2}{4a}$.
Since $a < 0$, the parabola opens downwards, and the vertex represents the maximum point.
Therefore, the maximum value of $y$ occurs at $x = -\frac{b}{2a}$.