{"id":"cb_question_bank:3cf63674-2d21-44e3-8f52-320185b9ecf4","source_id":"cb_question_bank","source_item_id":"3cf63674-2d21-44e3-8f52-320185b9ecf4","external_id":"fcee5450-b85d-43d4-a5c2-befadc67d6b0","ibn":null,"question_id":"2c05d312","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":"mcq","answer_source":"key","layout_warning":null,"form_id":null,"module_number":null,"position":null,"stem":"57 x squared plus left parenthesis 57 b plus a right parenthesis x plus a b equals 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 ?","stimulus":"","stimulus_html":null,"stem_html":"<p style=\"text-align: center;\"><math alttext=\"57 x squared plus left parenthesis 57 b plus a right parenthesis x plus a b equals 0\"><mrow><mn>57</mn></mrow><msup><mi>x</mi><mn>2</mn></msup><mo>+</mo><mo>(</mo><mrow><mn>57</mn></mrow><mi>b</mi><mo>+</mo><mi>a</mi><mo>)</mo><mi>x</mi><mo>+</mo><mi>a</mi><mi>b</mi><mo>=</mo><mn>0</mn></math></p>\n<p style=\"text-align: left;\">In the given equation, <math alttext=\"a\"><mi>a</mi>\n</math> and <math alttext=\"b\"><mi>b</mi>\n</math> are positive constants. The product of the solutions to the given equation is <math alttext=\"k a b\"><mi>k</mi><mi>a</mi><mi>b</mi></math>, where <math alttext=\"k\"><mi>k</mi>\n</math> is a constant. What is the value of <math alttext=\"k\"><mi>k</mi>\n</math>?</p>","rationale_html":"<p>Choice A is correct. The left-hand side of the given equation is the expression <math alttext=\"57 x squared plus left parenthesis 57 b plus a right parenthesis x plus a b\"><mn>57</mn><msup><mi>x</mi><mn>2</mn></msup><mo>+</mo><mfenced><mrow><mn>57</mn><mi>b</mi><mo>+</mo><mi>a</mi></mrow></mfenced><mi>x</mi><mo>+</mo><mi>a</mi><mi>b</mi></math>. Applying the distributive property to this expression yields <math alttext=\"57 x squared plus 57 b x plus a x plus a b\"><mn>57</mn><msup><mi>x</mi><mn>2</mn></msup><mo>+</mo><mn>57</mn><mi>b</mi><mi>x</mi><mo>+</mo><mi>a</mi><mi>x</mi><mo>+</mo><mi>a</mi><mi>b</mi></math>. Since the first two terms of this expression have a common factor of <math alttext=\"57 x\"><mrow>\n\t<mn>57</mn>\n\t<mi>x</mi>\n</mrow>\n</math> and the last two terms of this expression have a common factor of <math alttext=\"a\"><mi>a</mi>\n</math>, this expression can be rewritten as <math alttext=\"57 x left parenthesis x plus b right parenthesis plus a left parenthesis x plus b right parenthesis\"><mn>57</mn><mi>x</mi><mfenced><mrow><mi>x</mi><mo>+</mo><mi>b</mi></mrow></mfenced><mo>+</mo><mi>a</mi><mfenced><mrow><mi>x</mi><mo>+</mo><mi>b</mi></mrow></mfenced></math>. Since the two terms of this expression have a common factor of <math alttext=\"left parenthesis x plus b right parenthesis\"><mfenced><mrow><mi>x</mi><mo>+</mo><mi>b</mi></mrow></mfenced></math>, it can be rewritten as <math alttext=\"left parenthesis x plus b right parenthesis left parenthesis 57 x plus a right parenthesis\"><mfenced><mrow><mi>x</mi><mo>+</mo><mi>b</mi></mrow></mfenced><mfenced><mrow><mn>57</mn><mi>x</mi><mo>+</mo><mi>a</mi></mrow></mfenced></math>. Therefore, the given equation can be rewritten as <math alttext=\"left parenthesis x plus b right parenthesis left parenthesis 57 x plus a right parenthesis equals 0\"><mfenced><mrow><mi>x</mi><mo>+</mo><mi>b</mi></mrow></mfenced><mfenced><mrow><mn>57</mn><mi>x</mi><mo>+</mo><mi>a</mi></mrow></mfenced><mo>=</mo><mn>0</mn></math>. By the zero product property, it follows that&nbsp;<math alttext=\"x plus b equals 0\"><mi>x</mi><mo>+</mo><mi>b</mi><mo>=</mo><mn>0</mn></math> or <math alttext=\"57 x plus a equals 0\"><mn>57</mn><mi>x</mi><mo>+</mo><mi>a</mi><mo>=</mo><mn>0</mn></math>. Subtracting <math alttext=\"b\"><mi>b</mi>\n</math> from both sides of the equation&nbsp;<math alttext=\"x plus b equals 0\"><mi>x</mi><mo>+</mo><mi>b</mi><mo>=</mo><mn>0</mn></math> yields <math alttext=\"x equals negative b\"><mi>x</mi><mo>=</mo><mo>-</mo><mi>b</mi></math>. Subtracting <math alttext=\"a\"><mi>a</mi>\n</math> from both sides of the equation&nbsp;<math alttext=\"57 x plus a equals 0\"><mn>57</mn><mi>x</mi><mo>+</mo><mi>a</mi><mo>=</mo><mn>0</mn></math> yields <math alttext=\"57 x equals negative a\"><mn>57</mn><mi>x</mi><mo>=</mo><mo>-</mo><mi>a</mi></math>. Dividing both sides of this equation by <math alttext=\"57\"><mn>57</mn>\n</math> yields <math alttext=\"x equals StartFraction negative a Over 57 EndFraction\"><mi>x</mi><mo>=</mo><mfrac><mrow><mo>-</mo><mi>a</mi></mrow><mn>57</mn></mfrac></math>. Therefore, the solutions to the given equation are <math alttext=\"negative b\"><mrow>\n\t<mo>-</mo>\n\t<mi>b</mi>\n</mrow>\n</math> and <math alttext=\"StartFraction negative a Over 57 EndFraction\"><mfrac><mrow><mo>-</mo><mi>a</mi></mrow><mn>57</mn></mfrac></math>. It follows that the product of the solutions of the given equation is&nbsp;<math alttext=\"left parenthesis negative b right parenthesis left parenthesis StartFraction negative a Over 57 EndFraction right parenthesis\"><mfenced><mrow><mo>-</mo><mi>b</mi></mrow></mfenced><mfenced><mfrac><mrow><mo>-</mo><mi>a</mi></mrow><mn>57</mn></mfrac></mfenced></math>, or <math alttext=\"StartFraction a b Over 57 EndFraction\"><mfrac><mrow><mi>a</mi><mi>b</mi></mrow><mn>57</mn></mfrac></math>. It’s given that the product of the solutions of the given equation is <math alttext=\"k a b\"><mi>k</mi><mi>a</mi><mi>b</mi></math>. It follows that <math alttext=\"StartFraction a b Over 57 EndFraction equals k a b\"><mfrac><mrow><mi>a</mi><mi>b</mi></mrow><mn>57</mn></mfrac><mo>=</mo><mi>k</mi><mi>a</mi><mi>b</mi></math>, which can also be written as <math alttext=\"a b left parenthesis one fifty seventh right parenthesis equals a b left parenthesis k right parenthesis\"><mi>a</mi><mi>b</mi><mfenced><mfrac><mn>1</mn><mn>57</mn></mfrac></mfenced><mo>=</mo><mi>a</mi><mi>b</mi><mfenced><mi>k</mi></mfenced></math>. It’s given that <math alttext=\"a\"><mi>a</mi>\n</math> and <math alttext=\"b\"><mi>b</mi>\n</math> are positive constants. Therefore, dividing both sides of the equation&nbsp;<math alttext=\"a b left parenthesis one fifty seventh right parenthesis equals a b left parenthesis k right parenthesis\"><mi>a</mi><mi>b</mi><mfenced><mfrac><mn>1</mn><mn>57</mn></mfrac></mfenced><mo>=</mo><mi>a</mi><mi>b</mi><mfenced><mi>k</mi></mfenced></math> by <math alttext=\"a b\"><mrow>\n\t<mi>a</mi>\n\t<mi>b</mi>\n</mrow>\n</math> yields <math alttext=\"one fifty seventh equals k\"><mfrac><mn>1</mn><mn>57</mn></mfrac><mo>=</mo><mi>k</mi></math>. Thus, the value of <math alttext=\"k\"><mi>k</mi>\n</math> is <math alttext=\"one fifty seventh\"><mfrac><mn>1</mn><mn>57</mn></mfrac></math>.</p>\n<p>Choice B is incorrect and may result from conceptual or calculation errors.</p>\n<p>Choice C is incorrect and may result from conceptual or calculation errors.</p>\n<p>Choice D is incorrect and may result from conceptual or calculation errors.</p>","correct_answer":["A"],"has_media":false,"has_mathml":true,"has_table":false,"source_created_at":"2023-08-02T20:25:59.822000+00:00","source_updated_at":"2023-08-02T20:25:59.822000+00:00","options":[{"label":"A","content_html":"<p><math alttext=\"one fifty seventh\"><mfrac>\n\t<mn>1</mn>\n\t<mn>57</mn>\n</mfrac>\n</math></p>","ord":0},{"label":"B","content_html":"<p><math alttext=\"one nineteenth\"><mfrac>\n\t<mn>1</mn>\n\t<mn>19</mn>\n</mfrac>\n</math></p>","ord":1},{"label":"C","content_html":"<p><math alttext=\"1\"><mn>1</mn>\n</math></p>","ord":2},{"label":"D","content_html":"<p><math alttext=\"57\"><mn>57</mn>\n</math></p>","ord":3}],"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}}