X2+(k+1)x+k =0 has root a and b find the value of k if a=2b

Hey kongsocecilia, good to see you again. Step 1: Identify the coefficients and apply Vieta's formulas. For a quadratic equation Ax^2 + Bx + C = 0, the sum of roots is -(B)/(A) and the product of roots is (C)/(A). Given the equation x^2 + (k+1)x + k = 0, we have A=1, B=(k+1), and C=k. The roots are a and b. Sum of roots: a+b = -(k+1) (*) Product of roots: ab = k (**) Step 2: Use the given condition a=2b to substitute into the formulas. Substitute a=2b into equation (*): 2b+b = -(k+1) 3b = -(k+1) (1) Substitute a=2b into equation (**): (2b)b = k 2b^2 = k (2) Step 3: Solve the system of equations for b and k. From equation (1), express b in terms of k: b = -(k+1)/(3) Substitute this expression for b into equation (2): 2(-(k+1)/(3))^2 = k 2((k+1)^2)/(9) = k 2(k^2 + 2k + 1) = 9k 2k^2 + 4k + 2 = 9k Step 4: Rearrange into a quadratic equation and solve for k. 2k^2 + 4k - 9k + 2 = 0 2k^2 - 5k + 2 = 0 Factor the quadratic equation: 2k^2 - 4k - k + 2 = 0 2k(k-2) - 1(k-2) = 0 (2k-1)(k-2) = 0 This gives two possible values for k: 2k-1 = 0 2k = 1 k = (1)/(2) k-2 = 0 k = 2 The values of k are (1)/(2) and 2. The value of k is (1)/(2) or 2. Send me the next one 📸