For the equation x2 + (2k+2)x + 2k+5 = 0, find the value(s) of k for which the roots are equal. If k>0, find the roots of the new equation whose roots are 2 extalpha + extalpha extbeta and 2 extbeta +

2. (i) (a) Step 1: Identify the coefficients of the quadratic equation x^2 + (2k+2)x + 2k+5 = 0. The coefficients are a=1, b=(2k+2), and c=(2k+5). Step 2: Apply the condition for equal roots. For the roots and to be equal ( = ), the discriminant () of the quadratic equation must be zero. = b^2 - 4ac = 0 (2k+2)^2 - 4(1)(2k+5) = 0 Step 3: Solve the equation for k. 4(k+1)^2 - 4(2k+5) = 0 Divide the entire equation by 4: (k+1)^2 - (2k+5) = 0 Expand the square: k^2 + 2k + 1 - 2k - 5 = 0 k^2 - 4 = 0 k^2 = 4 k = ± sqrt(4) k = 2 or -2 2. (i) (b) Step 1: Use the condition k > 0 to select the value of k. From part (a), the positive value for k is k=2. Substitute k=2 into the original quadratic equation to find its roots: x^2 + (2(2)+2)x + 2(2)+5 = 0 x^2 + (4+2)x + (4+5) = 0 x^2 + 6x + 9 = 0 This equation can be factored as (x+3)^2 = 0. So, the roots are = -3 and = -3. Step 2: Calculate the new roots. The new roots are given as r_1 = 2 + and r_2 = 2 + . Using = -3 and = -3: First, calculate = (-3)(-3) = 9. r_1 = 2(-3) + 9 = -6 + 9 = 3 r_2 = 2(-3) + 9 = -6 + 9 = 3 The new roots are 3 and 3. Step 3: Form the new quadratic equation. A quadratic equation with roots r_1 and r_2 is given by $x^2 - (r