OG详解-OG3 数学1 Q18

Choice B is correct. Let x be the first integer and let y be the second integer. If the first integer is 11 greater than twice the second integer, then x = 2y +11. If the product of the two integers is 546, then xy = 546. Substituting 2y +11 for x in this equation results in (2y +11)y = 546. Distributing the y to both terms in the parentheses results in 2y² +11y = 546. Subtracting 546 from both sides of this equation results in 2y² +11y -546 = 0. The left-hand side of this equation can be factored by finding two values whose product is 2(-546), or -1,092, and whose sum is 11. The two values whose product is -1,092 and whose sum is 11 are 39 and -28. Thus, the equation 2y² +11y -546 = 0 can be rewritten as 2y² +28y -39y -546 = 0, which is equivalent to 2y (y -14)+ 39(y -14)= 0, or (2y + 39)(y -14)= 0. By the zero product property, it follows that 2y + 39 = 0 and y -14 = 0. Subtracting 39 from both sides of the equation 2y + 39 = 0 yields 2y =-39. Dividing both sides of this equation by 2 yields y = - 39​2​. Since y is a positive integer, the value of y is not -39​2. Adding 14 to both sides of the equation y -14 = 0 yields y =14. Substituting 14 for y in the equation xy = 546 yields x (14)= 546. Dividing both sides of this equation by 14 results in x = 39. Therefore, the two integers are 14 and 39, so the smaller of the two integers is 14.
Choice A is incorrect and may result from conceptual or calculation errors.Choice C is incorrect. This is the larger of the two integers.Choice D is incorrect and may result from conceptual or calculation errors.