{"id":"cb_question_bank:c359b60e-4e8e-4f3e-a116-fa64c968bb2c","source_id":"cb_question_bank","source_item_id":"c359b60e-4e8e-4f3e-a116-fa64c968bb2c","external_id":"6bc7b975-d9d8-4ebf-a201-229c5335d826","ibn":null,"question_id":"5edc8c98","program":"SAT","module":"math","domain_code":"P","domain":"Advanced Math","skill_code":"P.B.","skill":"Nonlinear equations in one variable and systems of equations in two variables","difficulty":"H","score_band":7,"answer_type":"spr","answer_source":"key","layout_warning":null,"form_id":null,"module_number":null,"position":null,"stem":"64 x squared minus left parenthesis 16 a plus 4 b right parenthesis x plus a b equals 0 In the given equation, a and b are positive constants. The sum of the solutions to the given equation is k left parenthesis 4 a plus b right parenthesis , where k is a constant. What is the value of k ?","stimulus":"","stimulus_html":null,"stem_html":"

64x2-16a+4bx+ab=0

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In the given equation, a\n and b\n are positive constants. The sum of the solutions to the given equation is  k4a+b, where k\n is a constant. What is the value of k\n?

","rationale_html":"

The correct answer is \n\t1\n\t16\n\n. Let p\n and q\n represent the solutions to the given equation. Then, the given equation can be rewritten as 64x-px-q=0, or 64x2-64p+q+pq=0. Since this equation is equivalent to the given equation, it follows that -16a+4b=-64p+q. Dividing both sides of this equation by -64\n yields 16a+4b64=p+q, or 1164a+b=p+q. Therefore, the sum of the solutions to the given equation, p+q, is equal to 1164a+b. Since it's given that the sum of the solutions to the given equation is k4a+b, where k\n is a constant, it follows that \n\tk\n\t=\n\t\n\t\t1\n\t\t16\n\t\n\n. Note that 1/16, .0625, 0.062, and 0.063 are examples of ways to enter a correct answer.

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Alternate approach: The given equation can be rewritten as 64x2-44a+bx+ab=0, where a\n and b\n are positive constants. Dividing both sides of this equation by 4\n yields 16x2-4a+bx+ab4=0. The solutions for a quadratic equation in the form Ax2+Bx+C=0, where A\n, B\n, and C\n are constants, can be calculated using the quadratic formula, x=-B+B2-4AC2A and x=-B-B2-4AC2A. It follows that the sum of the solutions to a quadratic equation in the form \n\t\n\t\t\n\t\t\tA\n\t\t\t\n\t\t\t\tx\n\t\t\t\t2\n\t\t\t\n\t\t\n\t\t+\n\t\t\n\t\t\tB\n\t\t\tx\n\t\t\n\t\t+\n\t\tC\n\t\n\t=\n\t0\n\n is -B+B2-4AC2A+-B-B2-4AC2A, which can be rewritten as -B+-B+B2-4AC-B2-4AC2A, which is equivalent to -2B2A, or \n\t-\n\t\n\t\tB\n\t\tA\n\t\n\n. In the equation 16x2-4a+bx+ab4=0, \n\tA\n\t=\n\t16\n\n, B=-4a+b, and \n\tC\n\t=\n\t\n\t\t\n\t\t\t\n\t\t\t\ta\n\t\t\t\tb\n\t\t\t\n\t\t\t4\n\t\t\n\t\n\n. Substituting 16\n for A\n and -4a+b for B\n in \n\t-\n\t\n\t\tB\n\t\tA\n\t\n\n yields --4a+b16, which can be rewritten as 1164a+b. Thus, the sum of the solutions to the given equation is 1164a+b. Since it's given that the sum of the solutions to the given equation is k4a+b, where k\n is a constant, it follows that \n\tk\n\t=\n\t\n\t\t1\n\t\t16\n\t\n\n. 

","correct_answer":[".0625","1/16"],"has_media":false,"has_mathml":true,"has_table":false,"source_created_at":"2023-08-02T20:25:59.823000+00:00","source_updated_at":"2023-08-02T20:25:59.823000+00:00","options":[],"attribution":{"source_id":"cb_question_bank","source_name":"College Board digital SAT question bank (community dump)","rights_holder":"College Board","canonical_url":"https://satsuitequestionbank.collegeboard.org/","retrieved_from":"https://raw.githubusercontent.com/mdn522/sat-question-bank/main/data/cb-digital-questions.json","attribution":"SAT practice questions © College Board, from the official SAT Suite Question Bank. Retrieved via the community dump at github.com/mdn522/sat-question-bank.","disclaimer":"SAT® and PSAT/NMSQT® are trademarks registered by the College Board, which is not affiliated with, does not sponsor, and does not endorse this project.","license":"College Board copyright; source repository carries no licence. Local use only.","redistributable":false,"item_count":2017}}