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# How to find the quadratic equation when given two solutions?

If the quadratic equation x2 + px + q = 0 has two solutions x1 and x2, then x1 + x2 = - p and x1 x2 = q. Note: p is the coefficient of the x term of the quadratic equation and q is the constant term of the quadratic equation. For our case, x1 = 2 and x2 = -1/2. Then x1 + x2 = 2 + (-1/2) = 4/2 - 1/2 = 3/2. x1 x2 = 2 × (-1/2) = -1. Then the quadratic function is: x2 - (x1 + x2)x + x1 x2 = 0. Substitute values into this quadratic equation, we get x2 - (3/2)x + (-1) = 0. Multiply 2 to each term of the equation, we get 2x2 - 3x - 2 = 0. This is the quadratic equation that the question asked.

To check the result, we need to solve the quadratic equation to see whether the solution of 2x2 - 3x - 2 = 0 satisfy the given values. For the quadratic equation 2x2 - 3x - 2 = 0, a = 2, b = -3 and c = -2. Then b2 - 4ac = (-3)2 - 4(2)((-2) = 9 + 16 = 25 = 52. Then x1 = [-b + square root of (b2 - 4ac)]/2a = [3 + 5]/[2 × 2] = 8/4 = 2. x2 = [-b - square root of (b2 - 4ac)]/2a = [3 - 5]/[2 × 2] = -2/4 = -1/2. So, its two solutions are 2 and -1/2, which satisfy the given condition. So, the quadratic equation is the solution.