OG详解-OG4 数学1 Q24

Choice C is correct. It’s given that the graphs of the equations in the given system intersect at exactly one point, (x, y), in the xy-plane. Therefore, (x, y) is the only solution to the given system of equations. The given system of equations can be solved by subtracting the second equation, y = 3x + a, from the first equation, y = 2x² -21x +64. This yields y - y =(2x² -21x + 64) - (3x + a), or 0 = 2x² -24x + 64-a. Since the given system has only one solution, this equation has only one solution. A quadratic equation in the form rx² + sx +t = 0, where r , s, and t are constants, has one solution if and only if the discriminant, s² -4rt , is equal to zero. Substituting 2 for r , -24 for s, and -a +64 for t in the expression s² -4rt yields (-24)² -(4)(2)(64-a). Setting this expression equal to zero yields (-24)² -(4)(2)(64-a)= 0, or 8a +64 = 0. Subtracting 64 from both sides of this equation yields 8a =-64. Dividing both sides of this equation by 8 yields a =-8. Substituting -8 for a in the equation 0 = 2x² -24x +64-a yields 0 = 2x² -24x +64 + 8, or 0 = 2x² -24x + 72. Factoring 2 from the right-hand side of this equation yields 0 = 2(x² -12x + 36). Dividing both sides of this equation by 2 yields 0 = x² -12x + 36, which is equivalent to 0 =(x -6)(x -6), or 0 =(x -6)². Taking the square root of both sides of this equation yields 0 = x-6. Adding 6 to both sides of this equation yields x = 6.
Choice A is incorrect. This is the value of a, not x. Choice B is incorrect and may result from conceptual or calculation errors. Choice D is incorrect and may result from conceptual or calculation errors.