Question #2c05d312 - SAT Suite Question Bank

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$57 x^{2} + (57 b + a) x + a b = 0$

In the given equation, $a$ and $b$ are positive constants. The product of the solutions to the given equation is $k a b$ , where $k$ is a constant. What is the value of $k$ ?

$\frac{1}{57}$
$\frac{1}{19}$
$1$
$57$

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Correct Answer: A

Rationale

Choice A is correct. The left-hand side of the given equation is the expression $57 x^{2} + (57 b + a) x + a b$ . Applying the distributive property to this expression yields $57 x^{2} + 57 b x + a x + a b$ . Since the first two terms of this expression have a common factor of $57 x$ and the last two terms of this expression have a common factor of $a$ , this expression can be rewritten as $57 x (x + b) + a (x + b)$ . Since the two terms of this expression have a common factor of $(x + b)$ , it can be rewritten as $(x + b) (57 x + a)$ . Therefore, the given equation can be rewritten as $(x + b) (57 x + a) = 0$ . By the zero product property, it follows that $x + b = 0$ or $57 x + a = 0$ . Subtracting $b$ from both sides of the equation $x + b = 0$ yields $x = - b$ . Subtracting $a$ from both sides of the equation $57 x + a = 0$ yields $57 x = - a$ . Dividing both sides of this equation by $57$ yields $x = \frac{- a}{57}$ . Therefore, the solutions to the given equation are $- b$ and $\frac{- a}{57}$ . It follows that the product of the solutions of the given equation is $(- b) (\frac{- a}{57})$ , or $StartFraction a b Over 57 EndFraction$ . It’s given that the product of the solutions of the given equation is $k a b$ . It follows that $\frac{a b}{57} = k a b$ , which can also be written as $a b (\frac{1}{57}) = a b (k)$ . It’s given that $a$ and $b$ are positive constants. Therefore, dividing both sides of the equation $a b (\frac{1}{57}) = a b (k)$ by $a b$ yields $\frac{1}{57} = k$ . Thus, the value of $k$ is $\frac{1}{57}$ .

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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