5. If $\alpha$ and $\beta$ are the roots of the equation $3x^2 - 2x - 1 = 0$, find the equation whose roots are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$. Step 1: Find the sum and product of roots for the original equation. 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: $$ \alpha\beta = \frac{c}{a} = \frac{-1}{3} $$ Step 2: Find the sum of the new roots. Let the new roots be $r_1 = \frac{\alpha}{\beta}$ and $r_2 = \frac{\beta}{\alpha}$. The sum of the new roots is: $$ S = r_1 + r_2 = \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} $$ We know that $\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$. $$ S = \frac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta} $$ Substitute the values from Step 1: $$ S = \frac{\left(\frac{2}{3}\right)^2 - 2\left(-\frac{1}{3}\right)}{-\frac{1}{3}} = \frac{\frac{4}{9} + \frac{2}{3}}{-\frac{1}{3}} $$ $$ S = \frac{\frac{4}{9} + \frac{6}{9}}{-\frac{1}{3}} = \frac{\frac{10}{9}}{-\frac{1}{3}} $$ $$ S = \frac{10}{9} \times (-3) = -\frac{10}{3} $$ Step 3: Find the product of the new roots. The product of the new roots is: $$ P = r_1 r_2 = \left(\frac{\alpha}{\beta}\right) \left(\frac{\beta}{\alpha}\right) = 1 $$ Step 4: Form the new quadratic equation. A quadratic equation with roots $r_1$ and $r_2$ is given by $x^2 - Sx + P = 0$. Substitute the values of $S$ and $P$: $$ x^2 - \left(-\frac{10}{3}\right)x + 1 = 0 $$ $$ x^2 + \frac{10}{3}x + 1 = 0 $$ Multiply the entire equation by $3$ to clear the denominator: $$ 3x^2 + 10x + 3 = 0 $$ The equation is $\boxed{3x^2 + 10x + 3 = 0}$. 6. Without solving the equation, find the sum and product of the roots of $3x^2 + 5x - 2 = 0$. Step 1: Identify coefficients. For a quadratic equation $ax^2 + bx + c = 0$, the sum of roots is $-\frac{b}{a}$ and the product of roots is $\frac{c}{a}$. In $3x^2 + 5x - 2 = 0$, we have $a=3$, $b=5$, $c=-2$. Step 2: Calculate the sum of the roots. $$ \text{Sum of roots} = -\frac{b}{a} = -\frac{5}{3} $$ Step 3: Calculate the product of the roots. $$ \text{Product of roots} = \frac{c}{a} = \frac{-2}{3} $$ 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. Step 1: Set up the equation. Let the first integer be $n$. The next consecutive integer is $n+1$. Their product is $n(n+1) = 132$. Step 2: Solve the quadratic equation. $$ n^2 + n = 132 $$ $$ n^2 + n - 132 = 0 $$ Factor the quadratic equation: $$ (n+12)(n-11) = 0 $$ This gives two possible values for $n$: $$ n = -12 \quad \text{or} \quad n = 11 $$ Step 3: Determine the pairs of consecutive integers. 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. Step 1: Define variables. Let the width of the rectangle be $w$ units. The length $l$ exceeds the width by 5 units, so $l = w+5$ units. Step 2: Set up the area equation. The area of a rectangle is given by $A = l \times w$. $$ (w+5)w = 104 $$ Step 3: Solve the quadratic equation. $$ w^2 + 5w = 104 $$ $$ w^2 + 5w - 104 = 0 $$ Factor the quadratic equation: $$ (w+13)(w-8) = 0 $$ This gives two possible values for $w$: $$ w = -13 \quad \text{or} \quad w = 8 $$ Step 4: Determine the valid dimensions. Since the width of a rectangle cannot be negative, we discard $w=-13$. Therefore, the width is $w=8$ units. The length is $l = w+5 = 8+5 = 13$ 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 $x = -\frac{b}{2a}$. Step 1: Understand the shape of the parabola. The equation $y = ax^2 + bx + c$ represents a parabola. The sign of the coefficient $a$ determines the direction in which the parabola opens. If $a > 0$, the parabola opens upwards, and its vertex is a minimum point. If $a < 0$, the parabola opens downwards, and its vertex is a maximum point. Step 2: Identify the vertex formula. The x-coordinate of the vertex of a parabola given by $y = ax^2 + bx + c$ is always $x = -\frac{b}{2a}$. Step 3: Conclude based on the conditions. Given that $a < 0$, the parabola opens downwards. This means the highest point on the graph is the vertex. Therefore, the function $y = ax^2 + bx + c$ attains its maximum value at the x-coordinate of its vertex, which is $x = -\frac{b}{2a}$.