For what values of j does the equation (2x +7)(x - 4) = -30 + jx have exactly one real solution? Express your answer as a list of numbers, separated by commas.

Answer 1

For the equation to have exactly one real solution, the discriminant must be equal to zero. The discriminant for a quadratic equation in the form ax^2 + bx + c = 0 is given by b^2 - 4ac.

In our equation (2x + 7)(x - 4) = -30 + jx, we can expand the left side to get 2x^2 - 8x + 7x - 28 = -30 + jx. Simplifying further, we get 2x^2 - x - 58 = -30 + jx.

Comparing this equation with the standard form ax^2 + bx + c = 0, we have a = 2, b = -1, and c = -58. The discriminant is then calculated as (-1)^2 - 4(2)(-58) = 1 + 464 = 465.

For the equation to have exactly one real solution, the discriminant should be zero. Since the discriminant is not zero for any real number j, the equation does not have exactly one real solution. So, the values of j for which the equation has exactly one real solution is an empty set: {}.