Question #6011a3f8 - SAT Suite Question Bank

Modded SAT Question Bank by Abdullah Mallik

$64 x^{2} + b x + 25 = 0$

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?

$negative 91$
$negative 80$
$5$
$40$

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Correct Answer: A

Rationale

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 $b squared minus 4 a c$ , is greater than 0, the given equation has exactly two real solutions. In the given equation, $64 x^{2} + b x + 25 = 0$ , $a equals 64$ and $c equals 25$ . Therefore, the given equation has exactly two real solutions when $left parenthesis b right parenthesis squared minus 4 left parenthesis 64 right parenthesis left parenthesis 25 right parenthesis greater than 0$ , or $b squared minus 6,400 greater than 0$ . Adding $6,400$ to both sides of this inequality yields $b squared greater than 6,400$ . Taking the square root of both sides of $b squared greater than 6,400$ yields two possible inequalities: $b less than negative 80$ or $b greater than 80$ . Of the choices, only choice A satisfies $b less than negative 80$ or $b greater than 80$ .

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.

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