64 x 2 + b x + 25 = 0 In the given equation, b is a constant. For which

Question

64 x 2 + b x + 25 = 0 In the given equation, b is a constant. For which of the following values of b will the equation have more than one real solution?

Accepted Answer

Choice A is correct. A quadratic equation of the form a x 2 + b x + c = 0 , where a , b , and c are constants, has either no real solutions, exactly one real solution, or exactly two real solutions. That is, for the given equation to have more than one real solution, it must have exactly two real solutions. When the value of the discriminant, or b 2 - 4 a c , is greater than 0, the given equation has exactly two real solutions. In the given equation, 64 x 2 + b x + 25 = 0 , a = 64 and c = 25 . Therefore, the given equation has exactly two real solutions when b 2 - 4 64 25 > 0 , or b 2 - 6,400 > 0 . Adding 6,400 to both sides of this inequality yields b 2 > 6,400 . Taking the square root of both sides of b 2 > 6,400 yields two possible inequalities: b < - 80 or b > 80 . Of the choices, only choice A satisfies b < - 80 or b > 80 . Choice B is incorrect and may result from conceptual or calculation errors. Choice C is incorrect and may result from conceptual or calculation errors. Choice D is incorrect and may result from conceptual or calculation errors.

Question 64-x-2