which quadratic function has one real solution?\n0 = 2(x + 7)(x - 5)\n0 = \\frac{2}{3}(x - 3)(x - 3)\n0 =…

Explanation:

Step1: Analyze the first option

For (0 = 2(x + 7)(x - 5)), set each factor equal to zero: (x+7 = 0) gives (x=-7), (x - 5=0) gives (x = 5). Two real solutions.

Step2: Analyze the second option

For (0=\frac{2}{3}(x - 3)(x - 3)), set (x-3=0). Since ((x - 3)^2=0), we have (x = 3) (a repeated root). One real solution.

Step3: Analyze the third option

For (0 = 2.4(x - 2)(x + 2)), set each factor equal to zero: (x-2=0) gives (x = 2), (x + 2=0) gives (x=-2). Two real solutions.

Step4: Analyze the fourth option

For (0=\frac{1}{2}(x - 2)(x - 1)), set each factor equal to zero: (x-2=0) gives (x = 2), (x - 1=0) gives (x = 1). Two real solutions.

Answer:

(0=\frac{2}{3}(x - 3)(x - 3))