Identify coefficients and Vieta's formulas for the given quadratic equation.

You're on a roll — Step 1: Identify coefficients and Vieta's formulas for the given quadratic equation. The given quadratic equation is 3x^2 - (p-4)x + (2p+1) = 0. Comparing this to Ax^2 + Bx + C = 0, we have: A = 3 B = -(p-4) = 4-p C = 2p+1 The sum of the roots ( + ) and the product of the roots ( ) are given by Vieta's formulas: + = -(B)/(A) = -(-(p-4))/(3) = (p-4)/(3) = (C)/(A) = (2p+1)/(3) Step 2: Use the given condition to find the value of p. The problem states that (+2)(+2) = 1. Expand this expression: + 2 + 2 + 4 = 1 + 2( + ) + 4 = 1 Substitute the expressions for + and from Step 1: (2p+1)/(3) + 2((p-4)/(3)) + 4 = 1 Multiply the entire equation by 3 to eliminate the denominators: 3((2p+1)/(3)) + 3((2(p-4))/(3)) + 3(4) = 3(1) (2p+1) + 2(p-4) + 12 = 3 2p+1 + 2p-8 + 12 = 3 Combine like terms: 4p + 5 = 3 4p = 3 - 5 4p = -2 p = -(2)/(4) p = -(1)/(2) a) The value of p is -(1)/(2). Step 3: Find the sum and product of the new roots. The new roots are (+2) and (+2). Sum of new roots: S' = (+2) + (+2) = + + 4 Substitute the value of p = -(1)/(2) into the expression for + : + = (p-4)/(3) = (-1)/(2)-43 = (-1)/(2)-(8)/(2)3 = (-9)/(2)3 = -(9)/(6) = -(3)/(2) Now calculate S': S' = -(3)/(2) + 4 = -(3)/(2) + (8)/(2) = (5)/(2) Product of new roots: P' = (+2)(+2) From the problem statement, we are given that (+2)(+2) = 1. So, P' = 1. Step 4: Form the new quadratic equation. A quadratic equation with roots r_1 and r_2 can be written as x^2 - (r_1+r_2)x + r_1r_2 = 0. Using S' = (5)/(2) and P' = 1: x^2 - ((5)/(2))x + 1 = 0 To eliminate the fraction, multiply the entire equation by 2: 2x^2 - 5x + 2 = 0 b) The equation whose roots are (+2) and (+2) is 2x^2 - 5x + 2 = 0. Got more? Send 'em!