AMC 12B 2025 (Problem 8)There are integers aaa and bbb such that the polynomial x3−5x2+ax+bx^3-5x^2+ax+bx3−5x2+ax+b has 4+54+\sqrt{5}4+5 as a root. What is a+ba+ba+b?(A) 13\text{(A)}\;13(A)13(B) 17\text{(B)}\;17(B)17(C) 20\text{(C)}\;20(C)20(D) 30\text{(D)}\;30(D)30(E) 68\text{(E)}\;68(E)68Solution 1Related TopicsCoreLinear Independence of Irrational NumbersMajorAlgebraic IdentitiesCheck AnswerYour answer:ABCDECheckHints (3)Hint 1(4+5)3−5(4+5)2+a(4+5)+b=0(4+\sqrt{5})^3-5(4+\sqrt{5})^2+a(4+\sqrt{5})+b=0(4+5)3−5(4+5)2+a(4+5)+b=0Hint 2(43+3⋅425+3⋅4(5)2+(5)3)−5(42+2⋅45+(5)2)+a(4+5)+b=0(4^3+3\cdot4^2\sqrt{5}+3\cdot4(\sqrt{5})^2+(\sqrt{5})^3)-5(4^2+2\cdot4\sqrt{5}+(\sqrt{5})^2)+a(4+\sqrt{5})+b=0(43+3⋅425+3⋅4(5)2+(5)3)−5(42+2⋅45+(5)2)+a(4+5)+b=0Hint 3(64+485+60+55)−5(16+85+5)+a(4+5)+b=0(64+48\sqrt{5}+60+5\sqrt{5})-5(16+8\sqrt{5}+5)+a(4+\sqrt{5})+b=0(64+485+60+55)−5(16+85+5)+a(4+5)+b=0(19+4a+b)+(13+a)5=0(19+4a+b)+(13+a)\sqrt{5}=0(19+4a+b)+(13+a)5=0⇒19+4a+b=0\Rightarrow 19+4a+b=0⇒19+4a+b=013+a=013+a=013+a=0⇒a=−13\Rightarrow a=-13⇒a=−13⇒19+4(−13)+b=0\Rightarrow 19+4(-13)+b=0⇒19+4(−13)+b=0⇒b=33\Rightarrow b=33⇒b=33Ans: a+b=−13+33=20\text{Ans: }a+b=-13+33=20Ans: a+b=−13+33=20Related Problems (7)AMC 10A 2020 (Problem 14)AMC 10B Spring 2021 (Problem 15)AMC 10B 2025 (Problem 7)AMC 10A 2012 (Problem 22)AMC 10B 2023 (Problem 14)BMO1 2016/2017 (Problem 3)AMC 12A 2022 (Problem 14)Solution 2Related TopicsCoreComplex Conjugates and Conjugate RootsMajorVieta's FormulaHints (4)Hint 14+5 is a root⇒4−5 is a root, too.4+\sqrt{5}\text{ is a root}\Rightarrow4-\sqrt{5}\text{ is a root, too.}4+5 is a root⇒4−5 is a root, too.Hint 2Suppose the third root is r.\text{Suppose the third root is }r.Suppose the third root is r.By Toolkit 30: Vieta’s Formula\text{By Toolkit 30: Vieta's Formula}By Toolkit 30: Vieta’s Formula(4+5)+(4−5)+r=−(−5)=5(4+\sqrt{5})+(4-\sqrt{5})+r=-(-5)=5(4+5)+(4−5)+r=−(−5)=58+r=58+r=58+r=5r=−3r=-3r=−3Hint 3By Toolkit 30: Vieta’s Formula\text{By Toolkit 30: Vieta's Formula}By Toolkit 30: Vieta’s Formula(4+5)(4−5)+(4+5)(−3)+(4−5)(−3)=a(4+\sqrt{5})(4-\sqrt{5})+(4+\sqrt{5})(-3)+(4-\sqrt{5})(-3)=a(4+5)(4−5)+(4+5)(−3)+(4−5)(−3)=a16−5−24=a⇒a=−1316-5-24=a\Rightarrow a=-1316−5−24=a⇒a=−13Hint 4By Toolkit 30: Vieta’s Formula\text{By Toolkit 30: Vieta's Formula}By Toolkit 30: Vieta’s Formula(4+5)(4−5)(−3)=−b(4+\sqrt{5})(4-\sqrt{5})(-3)=-b(4+5)(4−5)(−3)=−bb=(16−5)(3)=33b=(16-5)(3)=33b=(16−5)(3)=33Ans: a+b=−13+33=20\text{Ans: }a+b=-13+33=20Ans: a+b=−13+33=20Related Problems (9)AMC 10A/12A 2022 (Problem 16/15)AMC 10A/12A Spring 2021 (Problem 14/12)AMC 12B 2023 (Problem 14)AMC 12B 2022 (Problem 4)AMC 12B 2021 Spring (Problem 16)AMC 10A/12A 2025 (Problem 18/12)AMC 12A 2025 (Problem 19)AMC 12A 2024 (Problem 15)View all related problems →Final Answer(C) 202020