Given the quadratic equation 3x2 - 4x + 2 = 0, with roots alpha and beta.

Here's the solution to question 1: 1(i)a) Show that ^3 + ^3 = -(8)/(27) Given the quadratic equation 3x^2 - 4x + 2 = 0, with roots and . Step 1: Identify the coefficients and find the sum and product of roots. For a quadratic equation ax^2 + bx + c = 0, the sum of roots is + = -(b)/(a) and the product of roots is = (c)/(a). Here, a=3, b=-4, c=2. + = -(-4)/(3) = (4)/(3) = (2)/(3) Step 2: Use the identity for ^3 + ^3. The identity is ^3 + ^3 = ( + )(( + )^2 - 3 ). Step 3: Substitute the values of + and . ^3 + ^3 = ((4)/(3))(((4)/(3))^2 - 3((2)/(3))) ^3 + ^3 = ((4)/(3))((16)/(9) - 2) ^3 + ^3 = ((4)/(3))((16)/(9) - (18)/(9)) ^3 + ^3 = ((4)/(3))(-(2)/(9)) ^3 + ^3 = -(8)/(27) This shows the required result. 1(i)b) Find the equation with integral coefficients whose roots are (1)/(^2) and (1)/(^2) Let the new roots be r_1 = (1)/(^2) and r_2 = (1)/(^2). The general quadratic equation is x^2 - (sum of roots)x + (product of roots) = 0. Step 1: Calculate the sum of the new roots. S' = (1)/(^2) + (1)/(^2) = (^2 + ^2)/(( )^2) First, find ^2 + ^2: ^2 + ^2 = ( + )^2 - 2 Substitute the values from part (a): ^2 + ^2 = ((4)/(3))^2 - 2((2)/(3)) = (16)/(9) - (4)/(3) = (16)/(9) - (12)/(9) = (4)/(9) Now, calculate S': S' = (4)/(9)((2)/(3))^2 = (4)/(9)(4)/(9) = 1 Step 2: Calculate the product of the new roots. P' = ((1)/(^2))((1)/(^2)) = (1)/(( )^2) Substitute the value of : P' = (1)/((2)3)^2 = (1)/(4)9 = (9)/(4) Step 3: Form the new quadratic equation. The equation is x^2 - S'x + P' = 0. x^2 - 1x + (9)/(4) = 0 To obtain integral coefficients, multiply the entire equation by 4: 4x^2 - 4x + 9 = 0 The equation with integral coefficients is 4x^2 - 4x + 9 = 0. 1(ii) Find the set of real values of x for which |(x+3)/(x-1)| = (x+3)/(x-1) The property |A| = A holds true if and only if A 0. Therefore, we need to find the values of x for which (x+3)/(x-1) 0. Also, the denominator cannot be zero, so x-1 0 x 1. Step 1: Analyze the signs of the numerator and denominator. For the fraction to be non-negative, both numerator and denominator must have the same sign (or the numerator is zero). Case 1: Numerator is non-negative and denominator is positive. x+3 0 x -3 x-1 > 0 x > 1 The intersection of these conditions is x > 1. Case 2: Numerator is non-positive and denominator is negative. x+3 0 x -3 x-1 < 0 x < 1 The intersection of these conditions is x -3. Step 2: Combine the solutions from both cases. The set of real values of x is x -3 or x > 1. In interval notation, this is (-, -3] (1, ). The set of real values of x is x -3 or x > 1.