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Advanced Math / Nonlinear equations in one variable and systems of equations in two variables Difficulty: Hard

57x2+(57b+a)x+ab=0

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

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Explanation

Choice A is correct. The left-hand side of the given equation is the expression 57x2+57b+ax+ab. Applying the distributive property to this expression yields 57x2+57bx+ax+ab. 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 57xx+b+ax+b. Since the two terms of this expression have a common factor of x+b, it can be rewritten as x+b57x+a. Therefore, the given equation can be rewritten as x+b57x+a=0. By the zero product property, it follows that x+b=0 or 57x+a=0. Subtracting b from both sides of the equation x+b=0 yields x=-b. Subtracting a from both sides of the equation 57x+a=0 yields 57x=-a. Dividing both sides of this equation by 57 yields x=-a57. Therefore, the solutions to the given equation are - b and -a57. It follows that the product of the solutions of the given equation is -b-a57, or ab57. It’s given that the product of the solutions of the given equation is kab. It follows that ab57=kab, which can also be written as ab157=abk. It’s given that a and b are positive constants. Therefore, dividing both sides of the equation ab157=abk by a b yields 157=k. Thus, the value of k is 157.

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.