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).

5. If and are the roots of the equation 3x^2 - 2x - 1 = 0, find the equation whose roots are ()/() and ()/(). 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: + = -(b)/(a) = -(-2)/(3) = (2)/(3) Product of roots: = (c)/(a) = (-1)/(3) Step 2: Find the sum of the new roots. Let the new roots be r_1 = ()/() and r_2 = ()/(). The sum of the new roots is: S = r_1 + r_2 = ()/() + ()/() = (^2 + ^2)/() We know that ^2 + ^2 = (+)^2 - 2. S = ((+)^2 - 2)/() Substitute the values from Step 1: S = ((2)/(3))^2 - 2(-(1)/(3))-(1)/(3) = (4)/(9) + (2)/(3)-(1)/(3) S = (4)/(9) + (6)/(9)-(1)/(3) = (10)/(9)-(1)/(3) S = (10)/(9) × (-3) = -(10)/(3) Step 3: Find the product of the new roots. The product of the new roots is: P = r_1 r_2 = (()/()) (()/()) = 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 - (-(10)/(3))x + 1 = 0 x^2 + (10)/(3)x + 1 = 0 Multiply the entire equation by 3 to clear the denominator: 3x^2 + 10x + 3 = 0 The equation is 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 -(b)/(a) and the product of roots is (c)/(a). In 3x^2 + 5x - 2 = 0, we have a=3, b=5, c=-2. Step 2: Calculate the sum of the roots. Sum of roots = -(b)/(a) = -(5)/(3) Step 3: Calculate the product of the roots. Product of roots = (c)/(a) = (-2)/(3) The sum of the roots is -(5)/(3) and the product of the roots is -(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 or 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 11 and 12 or -12 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 × 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 or 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 width = 8 units, length = 13 units. 9. Show that if a < 0, y = ax^2 + bx + c has a maximum value when x = -(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 = -(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 = -(b)/(2a).