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 a quadratic equation $ax^2 + bx + c = 0$, the sum of roots is $\alpha + \beta = -\frac{b}{a}$ and the product of roots is $\alpha\beta = \frac{c}{a}$.
Here, $a=3$, $b=-2$, $c=-1$.
Sum of roots: $\alpha + \beta = -\frac{-2}{3} = \frac{2}{3}$.
Product of roots: $\alpha\beta = \frac{-1}{3}$.

The new roots are $r_1 = \frac{\alpha}{\beta}$ and $r_2 = \frac{\beta}{\alpha}$.

We know that $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$.
Substitute this into the expression for $S'$:

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

Find a common denominator for the numerator:

Multiply the entire equation by 3 to eliminate the fraction:

The equation whose roots are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$ 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 $= -\frac{b}{a} = -\frac{5}{3}$.

Product of roots $= \frac{c}{a} = \frac{-2}{3}$.

The sum of the roots is $\boxed{-\frac{5}{3}}$ and 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 $n+1$.

The product of the two consecutive integers is 132:

Factor the quadratic equation:

This gives two possible values for $n$:

If $n=11$, the integers are 11 and $11+1=12$.
Check: $11 \times 12 = 132$.
If $n=-12$, the integers are $-12$ and $-12+1=-11$.
Check: $(-12) \times (-11) = 132$.

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 exceeds its width by 5 units, so the length $l = w+5$ units.

The area of a rectangle is $l \times w$. Given the area is 104 square units.

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

This gives two possible values for $w$:

Since the width of a rectangle cannot be negative, we discard $w=-13$.
So, the width $w = 8$ units.
The length $l = w+5 = 8+5 = 13$ units.
Check: Area $= 13 \times 8 = 104$ square units.

The dimensions of the rectangle are length $\boxed{13 \text{ units}}$ and width $\boxed{8 \text{ units}}$.